A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.7 m from his television set. A reporter at the press conference is located 5.5 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.

Answer:

So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s

Answer:

Explanation:

purchase price = 400 tepizes / m²

1 tepiz = .625 dollar

purchase price in terms of dollar = 400 x .625 dollar / m²

= 250 dollar / m²

.9144 m = 1 yard

1 m = 1.0936 yard

1m² = 1.196 yard²

price in terms of dollar / yards²

= 250 / 1.196 dollar / yard²

= 209 dollar / yard²

Price of cloth in New York = 120 dollar / yard²

loss = 209 - 120 = 89 dollar / yard²

500 m² = 500 x 1.196 yard²

= 598 yard²

net loss in purchasing 500 m² cloth

= 598 x 89

= 53222 dollar .

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]

Answer:

Kinetic Energy of the disk = 252 J

Explanation:

weight of disk = 810 N

radius = 1.56 m

applied force = 49 N

time = 2.95 s

kinetic energy of disk = ?

first, we find the mass of the disk

mass of disk = weight/acceleration due to gravity(9.81 m/s^2) = 810/9.81 m/s^2

mass of disk = 82.57 kg

torque on the disk = force x radius = 49 x 1.56 = 76.44 N-m

moment of inertia I = m[tex]r^{2}[/tex] = 82.57 x [tex]1.56^{2}[/tex] = 200.9 kg-[tex]m^{2}[/tex]

recall that

Torque T = Iα

where α = angular acceleration

76.44 = 200.9α

α = 76.44/200.9 = 0.38 m/s^2

from the equation of angular motion,

ω = ω' + αt

where ω =  final angular speed

ω' = initial angular speed = 0 rad/s since disk starts from rest

t = time = 2.95 s

imputing values into the equation, we have

ω = 0 + (0.38 x 2.95)

ω = 1.12 rad/s

kinetic energy of the disk = I[tex]w^{2}[/tex]

KE = 200.9 x [tex]1.12^{2}[/tex]

Kinetic Energy of the disk = 252 J

Answer:

i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:

ii) Given:

Velocity = v

Time = t

Force = F

Force = mass x acceleration

         = mass x velocity/time

So, mass= (force x time) / velocity

[mass] = Ftv^-1

Hence, dimesnion of mass is Ftv^-1.

Answer:

the answer is opposite.

plz mark brainliest

Explanation:

Answer:

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

Explanation:

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]

Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]

Since, 1 m/s = 2.23 mph

4.28 m/s = 9.57 mph

So, the speed of the rider is 4.28 m/s or 9.57 mph.  

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

[tex]t_A=0.55s[/tex]

[tex]t_B=0.45s[/tex]

(a) The distance from the kicker to each of the 2 spectators is given by:

[tex]d_A=v \times t_A[/tex]

where,

v= speed of sound

[tex]t_A[/tex]=time taken for the sound waves to reach the ears

[tex]d_A=343\times 0.55=188.65[/tex]m

(b)[tex]d_B=v \times t_B[/tex]

where,

v= speed of sound

[tex]t_B[/tex]=time taken for the sound waves to reach the ears

[tex]d_B=343\times 0.45=154.35m[/tex]

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]

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 .

Answer:

I =1.8 kgm^2

Explanation:

In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:

[tex]\tau=I\alpha[/tex]        (1)

τ: torque applied to the door

I: moment of inertia of the door

α: angular acceleration = 5 rad/s^2

The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).

F = 10 N

d = 0.9 m

You replace the expression for τ, and solve for I:

[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]

The moment of inertia of the door is 1.8 kgm^2

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 .

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 .

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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Answer:

d = 41.91 m

Explanation:

In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:

For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:

[tex]x=x_o+v_1t[/tex]    (1)

xo: initial position of the front of the train = 500m

v1: speed of the train = 50km/h

For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):

[tex]x'=v_2t+\frac{1}{2}at^2[/tex]   (1)

v2: initial speed of superman = 100km/h

a: acceleration = 10m/s^2

When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:

[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]

Thus, you equal x=x'

[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]

You solve the last equation for t by using the quadratic formula:

[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]

You only use t1 = 3.02s because negative times do not have physical meaning.

Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):

[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]

x is the position of the front of the train, then, the dstance traveled by the train is:

d = 541.91m - 500m = 41.91 m

Answer:

100 J

Explanation:

Work = change in energy

W = ΔKE

W = ½ mv² − ½ mv₀²

W = ½ m (v² − v₀²)

W = ½ (10 kg) ((6 m/s)² − (4 m/s)²)

W = 100 J

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.

Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.

The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)

The horizontal distance traveled by particle is

x = ucosθt

t = x/ucosθ

The force in an electric field is F = qE...................(1)

where, q is charge , E is the strength of electric field

From, newton 2nd law of motion, Force F = ma.................(2)

Equating both the equations, we get

ma = qE

a = qE/m..................(3)

The vertical distance, y =usinθt - 1/2at²

From equation 3, we have

y = usinθt  -  1/2 (qE/m) t²

if y = 0, t = 2musinθ/(qE) = x / (ucosθ)

The electric field is represented as

Also, E = 2mu²×sinθ×cosθ/(xq)

Plug the values, we get

E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)

E = -4556.18 N/m

Thus, the electric field of the particle is  -4556.18 N/m.

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Answer:

The ratio is  [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

      Considering the second car

 The force causing it s movement  is mathematically represented as

       [tex]F_{T2} = ma[/tex]

 Considering the first car

 The force causing it s movement  is mathematically represented as

      [tex]F = F_{T1} -F_{T2} = ma[/tex]

=>   [tex]F_{T1} -ma = ma[/tex]

=>   [tex]F_{T1} = 2 ma[/tex]

=> [tex]\frac{F_{T1}}{ma} = 2[/tex]

=> [tex]\frac{F_{T1}}{F_{T2}} = 2[/tex]

Answer:

Mg or your weight.

Explanation:

When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.

Answer:

rt

Explanation:

Answer:D

Explanation:

Given

Same force is applied to each ball such that all have different masses

and Force is given by the product of mass and acceleration

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

[tex]a=\frac{F}{m}[/tex]

So acceleration of ball A

[tex]a_A=\frac{F}{0.5}=2F[/tex]

acceleration of ball B

[tex]a_B=\frac{F}{0.75}=\frac{4F}{3}=1.33F[/tex]

acceleration of ball C

[tex]a_C=\frac{F}{1}=F[/tex]

acceleration of ball D

[tex]a_D=\frac{F}{7.3}=\frac{F}{7.3}[/tex]

It is clear that acceleration of ball D is least.

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.

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

Answer:

a) 463.29 K

b) 8065.65 Pa

c) 0 J

Explanation:

The parameters given are;

Volume of the tank, V = 3.72 m³

Number of moles of gas present in the tank, n = 22.1 moles

Temperature of the gas before heating, T₁ = 300 k

Heat added to the gas, ΔQ = 4.5 × 10⁴ J

Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole

Avogadro's number = 6.022 × 10²³ particles per mole

a) ΔQ = n × [tex]c_v[/tex] × ΔT

Where:

ΔT = T₂ - T₁

T₂ = Final temperature of the gas

Hence, by plugging in the values, we have;

4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)

[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]

T₂ = 300 + 163.29 = 463.29 K

b) The pressure of the gas is found from the relation;

P×V = n×R×T

[tex]P = \dfrac{n \times R \times T}{V}[/tex]

Where:

P = Pressure of the gas

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature of the gas

V = Volume of the gas = 3.72 ³ (constant)

n = Number of moles of gas present = 22.1 moles (constant)

Hence the change in pressure is given by the relation;

[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]

Plugging in the values, we have;

[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]

c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;

W = f(P) × ΔV

The volume given in the question is constant

∴ ΔV = 0

Hence, W =  f(P) × 0 = 0 J

No work done by the gas during the process.

Answer:

Explanation:

  P = VI

  I = P/V = P/120

  R = V/I = V/(P/V) = V^2/P = 14400/P

Toaster: I = 1800/120 = 15 . . . amps

  R = 14400/1800 = 8 . . . ohms

Fry pan: I = 1300/120 = 10.833 . . . amps

  R = 14400/1300 = 11.08 . . . ohms

Lamp: I = 100/120 = 0.833 . . . amps

  R = 14400/100 = 144 . . . ohms

The total current exceeds 20 A, so will blow the fuse.

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

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.

Learn more about spring-mass system here:

https://brainly.com/question/16077243?referrer=searchResults

Explanation:

2.  No, not always.  Normal force is equal to force of gravity only when there's no acceleration in the vertical direction.

For example, when you stand in an elevator that's not moving, or moving at constant speed, then the normal force equals your weight.  But when the elevator accelerates upward, the normal force increases (making you feel heavier).  And when the elevator slows down, the normal force decreases (making you feel lighter).

3.  Yes, it is possible for an object to be moving eastward and experience a net force westward.  An example is a car applying the brakes.

4.  Friction force allows you to walk.  When you push against the floor, the floor's friction pushes back, as Newton's third law says.

If you try to walk on a slippery surface like ice, you won't be able to push against the ice, and the ice won't push back.

Answer:

Explanation:

At height h , potential energy of coaster car  having mass m = mgh .

The car will lose potential energy and gain kinetic energy.

height lost by car when it is at the top of loop of radius R

= h - 2R

potential energy lost = mg ( h - 2R )

kinetic energy gained = mg ( h - 2R )

kinetic energy = 0 + mg ( h - 2R )

= mg ( h - 2R )

b )

For the car to remain in contact with the track , if v be the minimum velocity

centripetal force at top = mg

m v² / R = mg

v² = gR

kinetic energy = 1/2 mv²

= 1/2 m x gR

= mgR /

If h be the minimum height that can give this kinetic energy

mg ( h - 2R ) = mgR / 2

h - 2R = R / 2

h = 2.5 R .

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

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)