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notsponge [240]
4 years ago
15

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st

art subscript, i, end subscript. Later, its speed increases to 4v_i4v i ​ 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Physics
1 answer:
OLEGan [10]4 years ago
8 0

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

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When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increase. The motion and spacing of the particles determines the state of matter of the substance. The end result of increased molecular motion is that the object expands and takes up more space.
4 0
3 years ago
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An ideal gas is compressed at constant pressure p=1500 Pa from a volume Vi=0.4 m3 to Vf=0.25 m3. At the same time, heat is trans
Anon25 [30]

To solve this problem we will apply the first law of thermodynamics and we will make a balance between the heat transferred, its internal energy and the total work. Recall that for gases the definition of work can be expressed in terms of its pressure and volume. Let's start

dQ = dU +dW

Here,

dU = Internal Energy

dW = Work

But internal energy is unchanged, then

dQ = dW

dQ = PdV

Where

dV= Change in Volume

P = Pressure

Finally, the expression of the heat transferred can be expressed in terms of pressure and volume, so it would end up becoming

dQ = p(v_i-v_f)

Replacing,

dQ = (1500)(0.4-0.25)

dQ = 225J

Therefore the correct answer is B.

3 0
4 years ago
A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude
Nadya [2.5K]

Answer:

<h2>a) S = \frac{1}{2}gt^2\\</h2><h2>b) 6secs</h2><h2>c) 192ft</h2>

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

S = ut + \frac{1}{2}at^{2}

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\

a) An expression for the altitude of the ballast after t seconds is therefore

S = \frac{1}{2}gt^2\\

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

7 0
3 years ago
What happens to the resistance of a wire as it gets wet
frozen [14]

It will cause corrosion of the wire inside

8 0
3 years ago
Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at
dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  F = 25 \  N

   The frequency is   f = 1 \  Hz

   The mass of the bridge per unit length is  \mu  =  2000 \  kg /m

    The length of the central span is  d =  144 m

     The oscillation amplitude of the section  considered at the time considered is  A = 75 \ mm =  0.075 \  m

      The time taken for the undriven oscillation to decay to \frac{1}{e}  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            m =  \mu  *  d

=>        m =  2000 * 144

=>        m =  288000 \ kg

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

                 A(t) = A_o e^{-\frac{bt}{2m} }

Here b is the damping constant and the A_o is the amplitude of the section when it was undriven

So from the question  

               \frac{A_o}{e}  = A_o e^{-\frac{b6T}{2m} }

=>            \frac{1}{e}  =e^{-\frac{b6T}{2m} }

=>          e^{-1} =e^{-\frac{b6T}{2m} }

=>           -\frac{3T b}{m}  =  -1

=>         b  = \frac{m}{3T}

Generally the amplitude of the section considered is mathematically represented as

           A =  \frac{n * F }{ b *  2 \pi }

=>       A =  \frac{n * F }{ \frac{m}{3T}  *  2 \pi }

=>       n =  A  *  \frac{m}{3}  *  \frac{2\pi}{25}

=>       n = 0.075 *  \frac{288000}{3}  *  \frac{2* 3.142 }{25}

=>       n = 1810 \ people

3 0
3 years ago
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