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2 years ago

A car travels at a speed of 40 m/s for 29.0 s;what is the distance traveled by the car?

Physics

1 answer:

lianna [129]2 years ago

8 0

Answer: 1160 m

Explanation:

Speed = distance / time. Plug in 40 m/s for speed and 29 s for time in order to get the distance, 1160 m.

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A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

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3 years ago

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3 years ago

When is the pressure of the man on the ground more – lying position or standing position? Explain.

Dmitrij [34]

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3 years ago

What do the key results indicate?

Artyom0805 [142]

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Explanation:

4 0

3 years ago

Read 2 more answers

A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released

statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

mass of the block, m = 2.4 kg
height of the block, h = 5 m
compression of the spring, x = 25 cm = 0.25 m

The spring constant is calculated as follows;

$F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m$

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

$\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2 - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2 = mgh - \frac{1}{2} kx^2\\\\mv^2 = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s$

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

8 0

3 years ago