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Ilia_Sergeevich [38]

4 years ago

15

it takes a school bus full of kids longer to stop than it does a small car with a single passenger. which law is this an example

o?

1 answer:

exis [7]4 years ago

7 0

Newton´s second law of motion

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A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

$L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}$

Where

$I_{i}\ and \ \omega_{i}$ initial moment of inertia and angular velocity

$I_{f}\ and \ \omega_{f}$ is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ is the satellite mass

r is the radus of the sphere

Substititute 1900kg for m and 4.6m for r

$I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}$

The final moment of inertia of the satellite about the centre of mass

$I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}$

Where $m_{x}$ is the antenna's mass and

I is the length of the antenna

$I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}$

So, the Final rotation rate of the satellite is:

$I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0

4 years ago

A 10 kg bicycle is traveling at a speed of 2 ms what is the bikes kinetic energy

natta225 [31]

The formula is KE=1/2MV^2

The kinetic energy of the bicycle is 40 J

5 0

3 years ago

A 410-g cylinder of brass is heated to 95.0*C and placed in a calorimeter containing 335 g of water at 25.0*C. The water is stir

yarga [219]

I don't know I am sorry

3 0

4 years ago

A body oscillates with 25hz what is its time period

Semmy [17]

Answer:

4 seconds

Explanation:

The frequency of a body is the number of oscillations in one second. It is the number of cycles per unit time. The S.I unit of frequency is the Hertz (Hz).

The period of a body is the time taken to complete one oscillation. The period is inversely proportional to the frequency of the body. It is the reciprocal of frequency and the S.I unit is second (s).

A body oscillates with 25hz. Therefore the frequency (f) = 25 Hz.

The period (T) is given as:

$Period (T)=\frac{1}{frequency(f)} \\T=\frac{1}{f} =\frac{1}{0.25}=4\\T=4\ seconds$

4 0

4 years ago

The work done to compress a spring with a force constant of 290.0 N/m a total of 12.3 mm is: a) 3.57 J b) 1.78 J c) 0.0219 J d)

iren2701 [21]

Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

$W=\dfrac{1}{2}kx^2$

$W=\dfrac{1}{2}\times 290\times (0.0123)^2$

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

7 0

4 years ago