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2 years ago

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A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2

.00 m/s^2. (a) How long does it take him to come to a stop? (b) How far does he travel as he is stopping?

1 answer:

Solnce55 [7]2 years ago

3 0

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so he travel 36 m to stop

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If an object is accelerating, it is traveling the same distance for each time interval of its motion

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Answer:

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Explanation:

if an object is accelerating the object will not travel the same distance every time interval

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A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

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a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

Learn more about work:

brainly.com/question/6763771

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3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

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Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

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3 years ago

When the displacement of a mass on a spring is 12 a the half of the amplitude, what fraction of the mechanical energy is kinetic

Sveta_85 [38]

You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).

And you know that the potential energy, PE, is [ 1/2 ] k (x^2)

Then, use x = A, to calculate the PE in the point where ME = PE.

ME = PE = [1/2] k (A)^2.

At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2

=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME

So, if PE is 1/4 of ME, KE is 3/4 of ME.

And the answer is 3/4

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3 years ago

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

<em>where</em>,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

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3 years ago