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2 years ago

A mimibus drives with a constant speed of 39 km/h. how far can it travel in 1.94 hours?

Physics

1 answer:

Rufina [12.5K]2 years ago

4 0

Answer:

<em>The minibus traveled 75.66 km</em>

Explanation:

<u>Motion with Constant Speed </u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

The formula to calculate the speed is:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we solve for d:

d = v . t

The minibus has a constant speed of v=39 km/h and it's required to find the distance it travels in t=1.94 hours.

Calculating the distance:

d = 39 km/h * 1.94 h

d = 75.66 km

The minibus traveled 75.66 km

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A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat

emmasim [6.3K]

Answer:

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Explanation:

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3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago

Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the

Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1 ⇒ w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

L₀ = L_f

I₀ w₀ = I w

w = $\frac{I_o}{I}$ w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

I <I₀

I₀/I > 1 ⇒ w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0

3 years ago

8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

3 0

2 years ago

The wasted energy in a charger is:

kirza4 [7]

Thermal is the wasted energy in a charger

4 0

3 years ago