uring a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? 26.7 N 16.7 N 13.3 N 107 N 40.0 N

Answers

Answer 1

Answer:

107 N, option d

Explanation:

Given that

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 20 m/s

final velocity of the ball, v = -12 m/s

time taken, Δt = 60 ms

Solving this question makes us remember "Impulse Theorem"

It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"

Mathematically, it is represented as

FΔt = m(v - u), where

F = the average force

Δt = time taken

m = mass of the ball

v = final velocity of the ball

u = initial velocity of the ball

From the question we were given, if we substitute the values in it, we have

F = ?

Δt = 60 ms = 0.06s

m = 0.2 kg

v = -12 m/s

u = 20 m/s

F = 0.2(-12 - 20) / 0.06

F = (0.2 * -32) / 0.06

F = -6.4 / 0.06

F = -106.7 N

Thus, the magnitude is 107 N


Related Questions

In which situation is chemical energy being converted to another form of energy?

Answers

Answer:

A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)

Explanation:

A person is swimming 1.1 m beneath the surface of the water in a swimming pool. A child standing on the diving board drops a ball into the pool directly above the swimmer. The swimmer sees the ball dropped from a height of 4.2 m above the water. From what height was the ball actually dropped?

Answers

Answer:

The actual height is  [tex]A =3.158 \ m[/tex]

Explanation:

From the question we are told that

   The depth of the person is  [tex]d = 1.1 \ m[/tex]

    The apparent height  is  [tex]D = 4.2 \ m[/tex]

Generally

     The refractive index of water is  [tex]n_w = 1.33[/tex]

      The refractive index of the air is  [tex]n_a = 1[/tex]

The apparent depth is mathematically represented as

      [tex]D = A [\frac{n_w}{n_a} ][/tex]  

substituting values

     [tex]4.2 = A [\frac{1.33}{1} ][/tex]  

=>   [tex]A = \frac{4.2 }{1.33}[/tex]

      [tex]A =3.158 \ m[/tex]

                 

The ball was dropped at the height of "3.158 m". To understand the calculation, check below.

Refractive Index

According to the question,

Water's refractive index, [tex]n_w[/tex] = 1.33

Air's refractive index, [tex]n_a[/tex] = 1

Apparent height, D = 4.2 m

Person's depth, d = 1.1 m

We know the relation,

→ D = A[[tex]\frac{n_w}{n_a}[/tex]]

By substituting the values, we get

4.2 = A[[tex]\frac{1.33}{1}[/tex]]

By applying cross-multiplication,

  A = [tex]\frac{4.2}{1.33}[/tex]

      = 3.158 m

Thus the approach above is correct.

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Gas is contained in a piston-cylinder assembly and undergoes three processes. First, the gas is compressed at a constant pressure of 100 [kPa] from initial volume of 1.0 [m3] to a volume of 0.5 [m3]. Second, the gas pressure is increased by heating at constant volume up to 200 [kPa]. Third, the gas is returned to its initial pressure and volume by a process for which P ∀=constant. All pressures given are absolute. For the gas as a system, is the system best considered open, closed, or isolated? Why?

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  B

Explanation:

The system is best considered a closed system because looking at process we can see that there was no exchange of matter between the system and the surrounding,(as the was no escape of matter from the system to the surrounding )

Secondly we can  deduce that there is  a variation in the volume. from [tex]1.0 m^3[/tex] to [tex]0.5 m^3[/tex]

   

A 1KW electric heater is switched on for ten minutes
How much heat does it produce?​

Answers

Explanation:

P=W/T ==> 1000w = Q/600 ==> Q=600000j

If a 1 - kilowatt electric heater is switched on for ten minutes then the heat produced by the electric heater would be 600 - kilo Joules .

What is thermal energy ?

It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy .

As given in the problem , we have to find out the heat produced by the 1 -  kilo watts electric heater if it is switched on for ten minutes ,

The heat produced by the electric heater = Power × time

                                                                      = 1000 × 600 Joules

                                                                      = 600 kilo - Joules

Thus , the heat produced by the electric heater would be 600 - kilo Joules .

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Question
20
what would be the advantages if your body had magnetic properties science subject​

Answers

Answer:

Some of the advantages if our body had magnetic properties are as follows:

Magnetic properties can have health benefits such as recovering quickly from a stroke, resolving bladder problems, and reducing blood pressure.Brain will be able to control more activities of the nervous system and other organs of the body using magnetic power.Heart will have many benefits of magnetic properties and able to provide more energy to the entire body through the circulation of blood.Magnetic properties in body will be able to maintain the production of melatonin that controls the sleep patterns.Magnetic properties will be able to kill cancer causing cells.

Hence, magnetic properties are somehow beneficial for humans.  

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

[tex]\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)[/tex]

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.)
(a) 1.00 cm
What is the general expression for the electric field along the axis of a uniformly charged ring? i MN/C
(b) 5.00 cm
i MN/C
(c) 30.0 cm
i MN/C
(d) 100 cm
i MN/C

Answers

Answer:

General Expression: E = kql/(l² + r²)^(3/2)

(a) 6.3 MN/C

(b) 22.8 MN/C

(c) 6.1 MN/C

(d) 0.63 MN/C

Explanation:

The general expression for electric field along axis of a uniformly charged ring is:

E = kqL/(L² + r²)^(3/2)

where,

E = Electric Field Strength = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q = Total Charge = 71 μC = 71 x 10⁻⁶ C

L = Distance from center on axis

r = radius of ring = 10 cm = 0.1 m

(a)

L = 1 cm = 0.01 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)

E = (6390 N.m³/C)/(0.00101 m³)

E =  6.3 x 10⁶ N/C = 6.3 MN/C

(b)

L = 5 cm = 0.05 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)

E = (31950 N.m³/C)/(0.00139 m³)

E =  22.8 x 10⁶ N/C = 27.4 MN/C

(c)

L = 30 cm = 0.3 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)

E = (191700 N.m³/C)/(0.03162 m³)

E =  6.1 x 10⁶ N/C = 6.1 MN/C

(d)

L = 100 cm = 1 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)

E = (639000 N.m³/C)/(1.015 m³)

E =  0.63 x 10⁶ N/C = 0.63 MN/C

John heats 1 kg of soup from 25 °C to 70 °C for 15 minutes by a heater. How long does the same heater take to heat 1.5 kg of the same kind of soup from 20 °C to 80 °C? The energy output per unit time by the heater is constant.

Answers

Answer:

30 minutes

Explanation:

Energy per time is constant, so:

E₁ / t₁ = E₂ / t₂

m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂

(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t

(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t

3/min = 90 / t

t = 30 min

Friction is a force that acts in an ___________ direction of movement.
a) similar
b) opposite
c) parallel
d) west

Answers

Answer:

the answer is opposite.

plz mark brainliest

Explanation:

A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magnetic field at a speed of 6.0 m/s. The magnetic field is perpendicular to the axis of the wire.

Required:
What is the angle between the magnetic field and the wire's velocity?

Answers

Answer:

Explanation: please see attached file I attached the answer to your question.

The angle between the magnetic field and the wire's velocity is 33.2 degrees.

Calculation of the angle:

Since the potential difference = 71mv = 71 *10 ^-3 V

The length is 12 cm = 0.12m

The magnetic field i.e. B = 0.27T

The speed or v = 4 m/s

here we assume [tex]\theta[/tex] be the angle

So,

e = Bvl sin[tex]\theta[/tex]

So,

[tex]Sin\theta[/tex] = e/bvl

= 71*10^-3 / 0.27 *4*0.12

= 0.5478

= 33.2 degrees

Therefore, the angle should be 33.2 degrees

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In each pair, select the body with more internal energy.

Answers

Answer:

rt

Explanation:

Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=

Answers

Answer:

Explanation shown below.

Explanation:

1.The number of images formed by 2 parallel mirrors is an infinite number of images.

2. The characteristics of a plane mirror is such that the object distance equals the image distance.

Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.

Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.

Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .

Hence the shortest distances are 8.7cm and 68.9cm

3. The number of reflections is infinite for both cases.

An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?

Answers

Speed = (distance) / (time)

Speed = (

Velocity = speed, and its direction

The velocity of the plane is 10.2 miles per second East.

(about 48 times the speed of sound)

The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one of the following frequencies will NOT resonate in the same pipe

a. 1800 Hz
b. 1000 Hz
c. 1400 Hz
d. 600 Hz
e. 400 Hz

Answers

Answer:

e. 400 Hz

Explanation:

In closed organ pipe,  only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz  , 1800 Hz .

Frequency not possible is 400 Hz .

Complete the first and second sentences, choosing the correct answer from the given ones.
1. The water temperature in the dish depends on the A / B / C / D.
A. average kinetic energy of water molecules
B. total kinetic energy of water molecules
C. water mass. D. potential energy of the container with water
2. The internal energy of the water in the vessel is E / F / G.
E. potential energy of the vessel with water
F. average kinetic energy of water molecules
G. sum of kinetic energy and potential water molecules

Answers

Answer:

Hope this helps :)

Explanation:

1. A

2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)

A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.

Answers

Answer:

Δm Δt> h ’/ 2c²

Explanation:

Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions

     ΔE Δt> h ’/ 2

     h’= h / 2π

to relate this to the masses let's use Einstein's relationship

      E = m c²

let's replace

     Δ (mc²) Δt> h '/ 2

the speed of light is a constant that we can condense exact, so

      Δm Δt> h ’/ 2c²

     

A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.

Answers

Answer:

  a) -5.26 m/s

  b) 27.91 m

Explanation:

a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.

  average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s

The change in velocity in the first half of the interval is ...

  Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s

So, the initial velocity (at the beginning of the last 1.85 s interval) is ...

  v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)

  v1 = -5.259 m/s

__

b) The velocity when the object hits the ground is ...

  v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s

This is related to the distance traveled by ...

  v² = 2dg . . . . . where g is the acceleration and d is the distance traveled

  d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters

The object travels a total distance of about 27.911 meters.

_____

The attached graph shows height vs. time.

someone please help me with this thanks

Answers

The dog has two legs

Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie

Answers

Answer:

The correct answer is C 1, 4, and 5 tie, then 2 and 3 tie

Explanation:

Solution

The electric field due to sheets E₁ positive =б/2E₀

E₂ is negative = б/2E₀

Now,

At the point 1, 4, 5 the electric field due to the sheets are in the opposite direction

At the point 1, the net field = -E₁ + E₂ =0

At the point A, the net field = -E₁ - E₂ = 0

Now,

At nay point inside between them, the electric field is seen to be at the same direction.

At the 2, 3 points the field is seen at the right

Thus,

E net = E₁ + E₂

= б/2E₀ + σ/2E₀

=б/E₀

Note: Kindly find an attached copy of the complete question to the solution

The correct answer is option C

The rank of the points according to the magnitude of the electric field is 1, 4, and 5 tie, then 2 and 3 tie

The magnitude of the electric field:

Let sheet 1 has positive surface charge density and sheet 2 has a negative surface charge density

The electric field (without direction) due to sheets will be

E₁ =σ/2E₀

E₂= σ/2E₀

Now,

At the point 1, 4, 5 the electric field due to the sheets is given by:

E = E₁ - E₂

E = σ/2E₀ - σ/2E₀

since the positive charge plate will have electric field lines away from the sheet and the negative charge plate will have electric field lines towards the sheet

E = 0

Now,

At points 2, 3 which are between the plates,

The net electric field is:

E = E₁ + E₂

since the electric field due to both the plates will be from positive to negative ( towards the negatively charged plate)

E = σ/2E₀ + σ/2E₀

E = σ/E₀

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The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and velocity vectors of the plane at a later time are given by r=(1.21x103i+3.45x104;)m and v= (2 i-3.5j) m/s The magnitude, in meters, of the plane's displacement from the origin is:_________
a. 2.50 x104
b. 1.45 x 104
c. 3.45x104
d. 2.5x103
e. none of the above

Answers

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

[tex]\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m[/tex]

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

[tex]d=\sqrt{(x-x_o)^2+(y-y_o)^2}[/tex]   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

[tex]d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m[/tex]

hence, the displacement of the airplane is 3.45*10^4 m

Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a static method of stretching it. You and your lab partner suspend the spring from a hook, hang different masses, m, on the lower end, and start them oscillating. One of you uses a meter stick to measure the amplitude, A, and the other uses a stopwatch to time 10 oscillations, t. Your data are as follows:Mass, m(g) Amplitude, A(cm) Time, T(s) 100 6.5 7.8150 5.5 9.8200 6.0 10.9250 3.5 12.4Use the best-fit line of an appropriate graph to determine the spring constant.

Answers

Answer:

  k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

         w = √ (k / m)

angular velocity and rel period are  related

         w = 2π / T

substitution

         T = 2π √(m / K)

in Experimental measurements give us the following data

  m (g)     A (cm)    t (s)   T (s)

  100        6.5         7.8    0.78

  150        5.5          9.8   0.98

   200      6.0        10.9    1.09

   250       3.5        12.4    1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

        T = t / 10

To find the spring constant we linearize the equation

        T² = (4π²/K)    m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

         m ’= 4π² / k

where m’ is the slope

           k = 4π² / m'

the equation of the line of the attached graph is

       T² = 0.00603 m + 0.0183

therefore the slope

       m ’= 0.00603  s²/g

    we calculate

         k = 4 π² / 0.00603

          k = 6547 g / s²

we reduce the mass to the SI system

         k = 6547 g / s² (1kg / 1000 g)

         k = 6,547 kg / s² =

         k = 6,547 N / m

let's reduce the uniqueness

         [N / m] = [(kg m / s²) m] = [kg / s²]

The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.

Given data:

Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].

Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].

Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].

The expression for the angular frequency of spring-mass system is,

[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex]  ......................................................(1)

Here, k is the spring constant.

Angular frequency is also expressed as,

[tex]\omega = 2 \pi f[/tex] .........................................................(2)

here, f is the linear frequency of spring-mass system.

And linear frequency is,

[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]

Then substitute equation (2) in equation (1) as,

[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]

Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.

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Escaping from a tomb raid gone wrong, Lara Croft (m = 62.0 kg) swings across an alligator-infested river from an 8.80-m-long vine. If her speed at the bottom of the swing is 6.30 m/s and she makes it safely across the river, what is the minimum breaking strength of the vine?

Answers

Answer:

887.2 N

Explanation:

Given that

Mass of Lara, m = 62 kg

Length of the vine, l = 8.8 m

Speed of the swing, v = 6.3 m/s Then,

We start by calculating her weight.

w = mass * acceleration

w = mg

w = 62 * 9.8

w = 607.6 N

F(c) = mv²/l, on substituting

F(c) = (62 * 6.3²) / 8.8

F(c) = 2460.78 / 8.8

F(c) = 279.6 N

T - 607.6 N = 279.6 N

T = 607.6 N + 279.6 N

T = 887.2 N

Thus, the minimum breaking strength of the vine is 887.2 N

The minimum breaking strength of the vine is about 900N.

Tension

Tension is a force developed by a rope, string, or cable when stretched under an applied force.

Given:

Mass (m) = 62 kg, length (l) = 8.8 m, velocity (v) = 6.3 m/s, g = 10 m/s².

Weight(W) = m * g = 62 * 10 = 620N

centripetal force = F(c) = mv²/l = 62 * 6.3² / 8.8 = 279.6N

T - W = F(c)

T - 620 = 279.6

T = 900N

The minimum breaking strength of the vine is about 900N.

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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 4.23 m/s, strikes the rod at angle θ = 46° from the normal at a distance D = 2/3 L, where L = 0.95 m, from the point of rotation and sticks to the rod after the collision.

Required:
What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?

Answers

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

calculate the volume of marble if its diameter is 10mm​

Answers

Answer:

The volume of the marble is [tex]523.33\ mm^2[/tex].

Explanation:

Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.

The volume of spherical shaped object is given by :

[tex]V=\dfrac{4}{3}\pi r^3[/tex]

Plugging all the values, we get :

[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]

So, the volume of the marble is [tex]523.33\ mm^2[/tex].

A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:_________

a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller, or equal to that observed inside the bus.
e. is greator than that observed inside the bus

Answers

Answer:

d

Explanation:

good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed

assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.

so a is correct

now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe

b is correct

assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus

e also correct

so correct answer is d. it depends on the speed of ball tossed by the student in front.

A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in the same direction, it would have hit the ground in:________.
a. 4.0 s.
b. 1.0 s.
c. 0.50 s.
d. 0.25 s.
e. 0.125 s.

Answers

Answer:

none of the answers is correct, the time  is the same  t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

     y = y₀ + I go t - ½ g t²

     0 = y₀ - ½ g t²

     t = √ 2y₀ / g

with the data from the first launch

     y₀i = ½ g t²

     y₀ = ½  9.8  0.6²

     y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

    t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time  is the same

t₁ = t₂ = 0.600 s

Astrophysicist Neil deGrasse Tyson steps into an elevator on the 29th floor of a skyscraper. For some odd reason, there is a scale on the floor of the elevator. Neil, whose mass is about 115 kg, decides to step on the scale and presses the button for a lower floor. The elevator starts traveling downwards with a constant acceleration of 1.5 m/s2 for 6.0 seconds, and then travels at a constant velocity for 6.0 seconds. Finally, the elevator has an upward acceleration of 1.5 m/s2 for 6.0 seconds as it comes to a stop.
A. If each floor is approximately 4 m tall, which floor does the elevator stop at?
B. If the mass of the elevator is 1,200 kg, what is the maximum tension of the elevator cable?

Answers

Answer:

A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)

B. T = 14859.5 N = 14.89 KN

Explanation:

A.

First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:

s₁ = Vi t + (0.5)at²

where,

s₁ = distance covered during accelerated downward motion = ?

Vi = initial speed = 0 m/s (since elevator is initially at rest)

t = time taken = 6 s

a = acceleration = 1.5 m/s²

Therefore,

s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²

s₁ = 4.5 m

also we find the final velocity using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (1.5 m/s²)(6 s)

Vf = 9 m/s

Now, the second part of downward motion is with constant velocity. So:

s₂ = vt

where,

s₂ = distance covered during constant speed downward motion = ?

v = Vf = 9 m/s

t = 6 s

Therefore,

s₂ = (9 m/s)(6 s)

s₂ = 54 m

Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:

s₃ = s₁ = 4.5 m

Therefore, the total distance covered by elevator is given by following equation:

s = s₁ + s₂ - s₃      (Downward motion taken positive)

s = 4.5 m + 54 m - 4.5 m

s = 54 m

Therefore, net motion of the elevator was 54 m downwards.

So the final floor will be:

Final Floor = Initial Floor - Distance Covered/Length of a floor

Final Floor = 29 - 54 m/4m

Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)

B.

The maximum tension will occur during the upward accelerated motion. It is given by the formula:

T = m(g + a)

where,

T = Max. Tension in Cable = ?

m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg

g = 9.8 m/s²

a = acceleration = 1.5 m/s²

Therefore,

T = (1315 kg)(9.8 m/s² + 1.5 m/s²)

T = 14859.5 N = 14.89 KN

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy is 91.19 GeV, but its short lifetime shows up as an intrinsic width of 2.5 GeV. what is the lifetime of this particle?

Answers

Answer:

The lifetime of the particle is  [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Explanation:

From the question we are told that

    The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]

    The intrinsic width is  [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]

The lifetime is mathematically represented as

     [tex]\Delta t = \frac{h}{\Delta E}[/tex]

Where h is the Planck's constant with a value of  [tex]1.055*10^{-34} \ J\cdot s[/tex]  

substituting values

    [tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]

     [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 35° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Part (a) Express the magnitude of the ball's initial horizontal velocity Or in terms of vo and 20%
Part (b) Express the magnitude of the ball's initial vertical velocity vOy in terms of vo and 0. 20%
Part (c) Find the ball's maximum vertical height Amat in meters above the ground.
Part (d) Create an expression in terms of vo-e, and g for the time-ur İt takes te ball to travel to its maximum vertical height.
Part (e) Calculate the horizontal distance in meters the ball has traveled when it returns to ground level.

Answers

Answer:

a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g,  y = 24.25 m

e) R = 138.46 m

Explanation:

This is a projectile launch exercise

a) let's use trigonometry to find the components of the initial velocity

  cos θ = v₀ₓ / v₀

  v₀ₓ = v₀ cos θ

   

v₀ₓ = 38 cos 35

v₀ₓ = 31.13 m / s

b) sin θ = [tex]v_{oy}[/tex] / v₀

    v_{oy} = v₀ sin θ

    v_{oy} = 38 sint 35

    v_{oy} = 21 80 m / s

c, d) to find the maximum height, the vertical speed is zero

     v_{y}² = v_{oy}² - 2 g y

     0 = [tex]v_{oy}[/tex]² - 2 gy

     y = v_{oy}² / 2g

let's calculate

     y = 21.80 2 / (2 9.8)

     y = 24.25 m

e) They ask to find the horizontal distance

    for this we can use the expression of reaches

       R = v₀² sin 2θ / g

let's calculate

      R = 38² sin (2 35) / 9.8

       R = 138.46 m

Question 9 of 10
2 Powie
You are riding a bicycle. You apply a forward force of 100 N, and you and the
bicycle have a combined mass of 80 kg. What is the acceleration of the
bicycle?
A. 125 m/s
B. 1.5 m/s2
c. 1.8 m/s?
D. 0.8 m/s​

Answers

Answer:

1.25 m/s^2

Explanation:

F = m*a ...... force = mass * acceleration

force = 100 N, mass = 80 kg

100 = 80 * a

100/80 = a = 1.25 m/s^2

Answer:

The acceleration is 1.25m/s².

Explanation:

You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :

[tex]f = m \times a[/tex]

Let F = 100,

Let m = 80,

[tex]100 = 80 \times a[/tex]

[tex]100 = 80a[/tex]

[tex]a = 100 \div 80[/tex]

[tex]a = 10 \div 8[/tex]

[tex]a = 1.25[/tex]

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