Answer:
v = 8[m/s]
Explanation:
In this case, the work that is exerted on the cart is the kinetic energy that allows the movement, so we can use the following equation of kinetic energy.
[tex]E_{kin}=Work\\E_{kin}=\frac{1}{2} *m*v^{2}[/tex]
Now replacing:
[tex]128=0.5*4*v^{2} \\v=\sqrt{64} \\v=8[m/s][/tex]
A force of 60 N is applied to a skier to pull him along a horizontal surface so that his speed remains constant. If the coefficient of friction of the skis on snow is 0.05, then what is the weight of the skier?
Answer:
[tex]1200\ \text{N}[/tex]
Explanation:
F = Force on the skier = 60 N
[tex]\mu[/tex] = Coefficient of friction = 0.05
w = Weight of skier
Force is given by
[tex]F=\mu w[/tex]
[tex]\Rightarrow w=\dfrac{F}{\mu}[/tex]
[tex]\Rightarrow w=\dfrac{60}{0.05}=\dfrac{6000}{5}[/tex]
[tex]\Rightarrow w=1200\ \text{N}[/tex]
Weight of the skier on which the force is being applied is [tex]1200\ \text{N}[/tex] .
A 1.40-kg ball bounces off a vertical wall. The ball approaches the wall at 9.70 m/s to the east and leaves the wall with the same speed. What is the change in momentum that the wall imparts to the ball
Answer:
The change in momentum is 0
Explanation:
Step one:
given data
mass of ball = 1.4kg
initial velocity of ball u = 9.7m/s
final velocity of ball v = 9.7m/s
Required
the change in momentum
Step two:
From the expression for momentum
P=mv
the change in momentum
Δp= mu-mv
Δp= 1.4*9.7-1.4*9.7
Δp= 13.58-13.58
Δp=0
There is no change in momentum
Objects 1 and 2 attract each other with a gravitational force
of 18.0 units. If the distance separating Objects 1 and 2 is
changed to one-third the original value, then the new
gravitational force will be units.
Answer:
F' = 162 units
Explanation:
The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:
[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}\\\\[/tex]
where,
F = gravitational force = 18 units
G = Gravitational Constant
m₁ = mass of object 1
m₂ = mass of object 2
r = distance between objects
Therefore,
[tex]18 = \frac{Gm_{1}m_{2}}{r^{2}}------ eqn (1)\\\\[/tex]
Now, if we change the value of distance to one-third of original value, then:
r' = r/3
[tex]F' = \frac{Gm_{1}m_{2}}{(\frac{r}{3})^{2}}\\\\F' = (9)(\frac{Gm_{1}m_{2}}{r^{2}})[/tex]
using eqn (1):
F' = 9(18 units)
F' = 162 units
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculate the force of gravity that the space shuttle experiences
Answer:
1.6675×10^-16N
Explanation:
The force of gravity that the space shuttle experiences is expressed as;
g = GM/r²
G is the gravitational constant
M is the mass = 1.0 x 10^5 kg
r is the altitude = 200km = 200,000m
Substitute into the formula
g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²
g = 6.67×10^-6/4×10^10
g = 1.6675×10^{-6-10}
g = 1.6675×10^-16N
Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N
A teacher asks students to make a model of a transform plate boundary the students use blocks to represent tectonic plates and Slide the blocks past each other in the directions of the arrows as shown which event can the students best demonstrate with their models
Answer:
Hello your question is incomplete hence I will give you a general answer as regards to tectonic plates sliding past each other in a sideways direction
answer : The Transform boundary is been demonstrated by the students when sliding tectonic plates past each other in sideways directions
Explanation:
The event that can be demonstrated by the students using blocks to represent tectonic plates and sliding the clocks past each other in sideways direction is Transform Boundary of the tectonic plates
A spring with a spring constant of 200 N/m stretches by 0.03 m. What is the potential energy of the spring?
Answer: 0.09 J
Explanation: K = 200 N/m , 1/2 X 200 N/m X (0.03 M)^2 = 0.09 J
the bouncing back of sound when it hits ahard surface is called
Answer:
Reflection of sound
Explanation:
Sound waves bounce back from hard surface's.
What is the relationship among the potential drops across each resistor when the resistors are in series
Answer:
V = V₁ + V₂ + V₃ + ... + Vₙ
Explanation:
When the resistors are connected end to end such that there is only one path for the current to follow, it is called a series arrangement of resistors. In the series arrangement of the resistors, the current across each resistor is the same as the current applied across the circuit.
The potential difference across each resistor is different in the series arrangement of the resistors. But the sum of potential differences across each of the resistors in the series arrangement of resistors is equal to the total potential difference applied by the battery or source. Therefore, if n number of resistors are connected in a series arrangement with a source of potential V, the:
V = V₁ + V₂ + V₃ + ... + Vₙ
a railway train 100m long passes over a bridge 400m long at 20ms.1 how long will it take to pass over the bridge
Total length to be covered=400+100=500mts
time =distance/speed
=500/20
=25 sec
3. Superman is flying on the sky, suddenly approaches an airplane. It is known that this airplane has 1,000,000 kg*m/s of momentum. Superman decides to push the airplane with a force of 1,200,000 N for .09 seconds. What is the impulse that superman gives to the plane? What is the new momentum of the airplane?
Answer:
Superman's delivered impulse : 108,000 kg m/s
New momentum of the airplane: 1,108,000 kg m/s
Explanation:
Recall that impulse can be estimated by multiplying the applied force times the duration of time the force was applied. Therefore, the impulse added by Superman was:
1,200,000 * 0.09 = 108,000 kg m/s
and then, the new momentum of the plane is the addition:
1000000 + 108000 = 1,108,000 kg m/s
whitch two options are forms of kinetic energy?
Answer:the witch has nothing to do with the problem
Explanation:
How many light years are there in one mile?
Answer:
1.70108e-13 , this is the answer hope it helps
a traveling wave is described by equation y(x,t)=0.003(20x+200t) whereby y and x are measured in meters and t in second .what is the period of this wave?
Answer:
0.0314secs
Explanation:
The standard equation of a wave is expressed as;
y(x,t) = Asin(2πx/λ+2πft)
compare and contrast with the equation y(x,t)=0.003(20x+200t)
2πft = 200t
2πf = 200
f = 200/2π
f = 100/π
Since period T = 1/f
T = π/100
T = 3.14/100
T = 0.0314secs
hence the period of the wave is 0.0314secs
Please help if giving 20 points and brainliest
You have designed a prototype of a new, lighter material. What should be the
next step in your process?
O A. Figure out what problems people want to solve.
B. Update your initial design for the new type of material.
O c. Test to see whether your material is strong and works well.
D. Brainstorm new ways to design materials.
Answer:
O
Explanation:
Because your product will not work well so people will not buy it and it could be a defect and explode
Answer:
c
Explanation:
what does newton's first law describes
Vector ~A has a negative x-component 3.07 units in length and a positive y-component 3.17 units in length. When a vector ~B = b1i + b2j is added to vector ~A the resulting vector has no x component and a negative y component of 4.43 units. Find the value of
a. b1
b. b2
Answer:
a. 3.07 b. 1.26
Explanation:
Given that A = -3.07i + 3.17j and B = b1i + b1j and C = A + B = 0i + 4.43j
Since A + B = -3.07i + 3.17j + b1i + b2j
= (-3.07 + b1)i + (3.17 + b2)j
So,(-3.07 + b1)i + (3.17 + b2)j = 0i + 4.43j
Comparing components,
-3.07 + b1 = 0 (1) and 3.17 + b2 = 4.43 (2)
a. From (1), b1 = 3.07
b. From(2) b2 = 4.43 - 3.17 = 1.26
Find the radioactivity of a 1 g sample of 226Ra given that [tex]t_{1/2}=1620[/tex] years and Avogadro's number = 6.023 × [tex]10^{23}[/tex].
Answer:
Explanation:
No of atoms of Ra in 1 g of sample = 6.023 x 10²³ / 226
N = 2.66 x 10²¹
disintegration constant λ = .693 / half life
half life = 1620 x 365 x 60 x 60 x 24 = 5.1 x 10¹⁰ s
disintegration constant λ = .693 / 5.1 x 10¹⁰
radioactivity dn / dt = λN
= (.693 / 5.1 x 10¹⁰ ) x 2.66 x 10²¹
= .3614 x 10¹¹ per sec
= 3.614 x 10¹⁰ / s
A student drops an object from rest above a force plate that records information about the force exerted on the object as a function of time during the time interval in which the object is in contact with the force plate. Which of the following measurements should the student take, in addition to the measurements from the force plate, to determine the change in momentum of the object from immediately before the collision to immediately after the collision?
a. The mass of the object.
b. The final speed of the object MOH 5000
c. The distance fallen by the object
d. The student has enough information to make the determination
Answer:
D
Explanation:
The student has enough information to make the determination
Measurements should the student take, in addition to the measurements from the force plate, to determine the change in momentum of the object from immediately before the collision to immediately after the collision the student has enough information to make the determination. Thus, option D is correct.
What happens during the experiment?A student drops an object from rest above a force plate that records information about the force exerted on the object as a function of time during the time interval in which the object is in contact with the force plate.
Momentum has the measure of motion of the object. Momentum is given by the product of mass and the velocity of the object. The law of conservation of momentum states that the momentum before the collision is equal to the momentum after the collision. It also states that the total momentum of a system or a body remains constant.
Therefore, Measurements should the student take, in addition to the measurements from the force plate, to determine the change in momentum of the object from immediately before the collision to immediately after the collision the student has enough information to make the determination. Thus, option D is correct.
Learn more about Measurements on:
https://brainly.com/question/2107310
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What is the electrical consumption in KVA of a motor powered by a 3-phase, 60 Hz, 460 VAC supply that continuously draws 17 A
Answer:
15.34 kVA
Explanation:
A motor is a device that converts electrical energy into mechanical energy. It takes in electrical energy at the input and produce torque (motion) at the output.
The power consumption for a three phase motor is the product of voltage and current and √3. The √3 is because it is a three phase supply.
Hence Power (P) =√3 × voltage (V) × current (I)
P = √3 × V × I
Given that voltage (V) = 460 V, current (I) = 17 A. Hence:
P = √3 × V × I = √3 × 460 × 17 = 13544.64 VA
But 1000 VA = 1 kVA. Hence:
[tex]P=13544.64\ VA*\frac{1\ kVA}{1000\ VA}=13.54\ kVA[/tex]
Light with a wavelength of 560.0 nm is incident on a pair of slits with a separation of 0.380 mm. (a) Find the angles corresponding to the locations of the first three orders of bright fringes away from the central bright fringe
Answer:
Angles corresponding to the locations of the first three orders of bright fringes away from the central bright fringe are;
∅₁ = 0.8439°
∅₂ = 0.1688°
∅₃ = 0.2533°
Explanation:
Given that;
wavelength λ = 560 nm = 560 × 10⁻⁹
Separation between slits d = 0.380 mm = 0.00038
n = first three orders = 1st order, 2nd order and 3rd oder.
we know that for constructive interference;
λn = dsin∅
sin∅ = λn/d
∅ = sin⁻¹ ( λn/d )
where λ is wavelength, ∅ is the angle, d is the distance between slits, n is the order of constructive interference.
now;
-First order; n = 1
∅₁ = sin⁻¹(λn/d) = sin⁻¹( (560 × 10⁻⁹)×(1) /0.00038 )
∅₁ = sin⁻¹( 0.001473) = 0.8439°
-2nd order; n = 2
∅₂ = sin⁻¹(λn/d) = sin⁻¹( (560 × 10⁻⁹)×(2) /0.00038 ) =
∅₂ = sin⁻¹( 0.002947) = 0.1688°
-3rd order; n = 3
∅₃ = sin⁻¹(λn/d) = sin⁻¹( (560 × 10⁻⁹)×(3) /0.00038 ) =
∅₃ = sin⁻¹( 0.004421) = 0.2533°
Therefore, angles corresponding to the locations of the first three orders of bright fringes away from the central bright fringe are;
∅₁ = 0.8439°
∅₂ = 0.1688°
∅₃ = 0.2533°
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5 m/s . It then flies a further distance of 46100 m , and afterwards, its velocity is 40.7 m/s . Find the airplane's acceleration.
Answer:
the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Explanation:
Given;
initial velocity of the airplane. u = 34.5 m/s
distance traveled by the airplane, s = 46,100 m
final velocity of the airplane, v = 40.7 m/s
The acceleration of the airplane is calculated from the following kinematic equation;
v² = u² + 2as
[tex]2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2[/tex]
Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Please help!! There’s 10 points for it
Answer:
1200 J
Explanation:
option 1 should be the answer
Do this for alot of points
1. A complete fitness and exercise program should incorporate three basic components: Endurance (Aerobic), Flexibility, and Strength. Each of these components has specific guidelines, which govern their effectiveness
1. Calculate the heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.750.000 joules of heat,
and its temperature changes from 32°C to 57°C.
Answer:
1.8 J/gºC
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 1500 g
Heat (Q) absorbed = 67500 J
Initial temperature (T₁) = 32 °C
Final temperature (T₂) = 57 °C
Specific heat capacity (C) =?
Next, we shall determine the change in temperature of the wood. This can be obtained as follow:
Initial temperature (T₁) = 32 °C
Final temperature (T₂) = 57 °C
Change in temperature (ΔT) =.?
ΔT = T₂ – T₁
ΔT = 57 – 32
ΔT = 25 °C
Finally, we shall determine the heat capacity of the wood. This can be obtained as follow:
Mass (M) = 1500 g
Heat (Q) absorbed = 67500 J
Change in temperature (ΔT) = 25 °C
Specific heat capacity (C) =?
Q = MCΔT
67500 = 1500 × C × 25
67500 = 37500 × C
Divide both side by 37500
C = 67500 / 37500
C = 1.8 J/gºC
Thus, the heat capacity of the wood is 1.8 J/gºC
The normal eye, myopic eye and old age
As we age, the lens of the eye hardens and accommodation no longer occurs: it is presbyopia.
Paradoxically, with age, a nearsighted person sees better than a "normal" person.
The aim of this exercise is to explain this paradox. The eye is modeled by a constant focal
length f′0 when the eye is presbyopic and by a screen (the retina) at the distance d from the
lens (d = 15 mm).
1. A normal, presbyopic eye sees an object at infinity in focus while a myopic eye sees an
object at a distance Dm from the eye (Dm = 15 cm).
What relation do we have in the two cases between f′0 and d?
2. A presbyopic person reads a newspaper placed 25 cm from his/her eyes. The radius r0 of
the pupil of the eye is 1 mm. Calculate the diameter of the spot on the retina image from a
log point, for a myopic eye and a normal eye. Conclude
Answer:
1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) the two diameters have the same order of magnitude and are very close to each other
Explanation:
You have some problems in the writing of your exercise, we will try to answer.
1) The equation to be used in geometric optics is the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where p and q are the distance to the object and the image, respectively, f is the focal length
* For the normal eye and with presbyopia
the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)
[tex]\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}[/tex]
f'₀ = 1.5 cm
this is the focal length for this type of eye
* Eye with myopia
the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm
1 / f = 1/15 + 1 / 1.5
1 / f = 0.733
f = 1.36 cm
this is the focal length for the myopic eye.
In general, the two focal lengths are related
f’₀ / f = 1.5 / 1.36
f’₀ / f = 1.10
The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit
a sin θ= m λ
the first zero occurs for m = 1, as the angles are very small
tan θ = y / f = sin θ / cos θ
for some very small the cosine is 1
sin θ = y / f
where f is the distance of the lens (eye)
y / f = lam / a
in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor
y / f = 1.22 λ / D
y = 1.22 λ f / D
where D is the diameter of the eye
D = 2R₀
D = 2 0.1
D = 0.2 cm
the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation
* normal eye
the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation
\frac{1}{f} = \frac{1}{p} + \frac{1}{q}
sustitute
[tex]\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}[/tex]
[tex]\frac{1}{f}[/tex]= 0.7066
f = 1.415 cm
therefore the diffraction is
y = 1.22 550 10⁻⁹ 1.415 / 0.2
y = 4.75 10⁻⁶ m
this is the radius, the diffraction diameter is
d = 2y
d_normal = 9.49 10⁻⁶ m
* myopic eye
In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm
[tex]\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}[/tex]
[tex]\frac{1}{f}[/tex]= 0.733
f = 1.36 cm
diffraction is
y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2
y = 4.56 10-6 m
the diffraction diameter is
d_myope = 2y
d_myope = 9.16 10-6 m
[tex]\frac{d_{normal}}{d_{myope}}[/tex] = 9.49 /9.16
\frac{d_{normal}}{d_{myope}} = 1.04
we can see that the two diameters have the same order of magnitude and are very close to each other
Two small nonconducting spheres have a total charge of (a) When placed 28.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive. What is the charge on each
Answer:
q = 1 x 10⁻⁵ C = 10 μC
Explanation:
The repulsive force between the charges is given by Coulumb's Law:
[tex]F = \frac{kq_{1}q_{2}}{r^{2}}\\[/tex]
where,
F = Electrostatic Force = 12 N
k = Coulomb's Constat = 9 x 10⁹ Nm²/C²
r = distance between charges = 28 cm = 0.28 m
Since the values or charges are not given. We assume that both charges have same mahnitude. Therefore,
q₁ = q₂ = q = charge on each sphere = ?
Therefore,
[tex]12\ N = \frac{(9\ x\ 10^{9}\ Nm^{2}/C^{2})q^{2}}{(0.28\ m)^{2}} \\\\q^{2} = \frac{(12\ N)(0.28\ m)^{2}}{9\ x\ 10^{9}\ Nm^{2}/C^{2}}\\q = \sqrt{1\ x\ 10^{-10}\ C^{2}}\\[/tex]
q = 1 x 10⁻⁵ C = 10 μC
Explain why a bolt of lightning is like the spark you might see when you touch a metal object and get a shock.
Answer:
Just like lightning, the spark you see is the discharge of static electricity that equalizes the charges. When you touch a metal object and get a shock, electrons are travelling in between you and the object to equalize the charges of the two objects. The light that is seen is the plasma created by electrons jumping between objects which heats the air surrounding them.
A small car with a mass of 800kg moving with a velocity of 27.8 m/s. The car stops at a yellow light in 3.9 seconds. What force did it take for the car to stop?
Answer:
F = 5702.56 N
Explanation:
Given that,
Mass of a small car, m = 800 kg
Initial speed of the car, u = 27.8 m/s
Final speed, v = 0
Time, t = 3.9 s
We need to find the force did it take for the car to stop.
The force acting on an object is given by :
[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{800\times (0-27.8)}{3.9}\\\\F=-5702.56\ N[/tex]
So, the magnitude of force acting on the car to stop is 5702.56 N.
A 15.0 kg bowling ball rolling at 3 m/s is stopped by a constant force of 11.2 newtons. Calculate the time the force must act on the bowling ball to stop it.
Answer:
4.01 seconds
Explanation:
Given that:
Mass of ball = 15kg
Initial velocity, u = 3m/s
Final velocity, v = 0
Force, F= 11.2 N
Change in velocity, dv = 3 - 0 = 3
Time force must act on the ball before stopping it:
Using the relation :
F = ma
a = (v - u) / t
Ft = m(v - u)
11.2 * t = 15 * 3
11.2t = 45
11.2t = 45
t = -+¯ 45 / 11.2
t = 4.01
t = 4 seconds.
PLEASE ANSWER 50 points
A car manufacturer wants to change its car’s design to increase the car’s acceleration. Which changes should the engineers consider making to the design?
O increase the force that the engine provides
O decrease the force that the engine provides
O increase the mass of the car
O decrease the mass of the car
O increase the top velocity the car can travel
O decrease the top velocity the car can travel
