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Readme [11.4K]
2 years ago
15

A train is moving with a velocity of 70 km/h. If north is taken as the positive direction, what is the direction of movement of

the train?
Physics
1 answer:
Anna11 [10]2 years ago
4 0

Answer:

+70 km/h north

Explanation:

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A 150 g egg is dropped from 3.0 meters. The egg is
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<u><em>Answer:</em></u>

<u><em> </em></u>

<u><em>9.2 N, with significant figure rounding (2 s.f.) </em></u>

<u><em></em></u>

<u><em>Explanation:</em></u>

<u><em>This problem can be solved using momentum. The following equation relates momentum (mass & velocity) with force and time:</em></u>

<u><em></em></u>

<u><em>Note that  where v is the final velocity and v₀ is the initial velocity. Δv just means change in velocity.</em></u>

<u><em></em></u>

<u><em>Mass of the egg is 150 g, but we need to convert to kilograms if we want to use Newtons as a unit. 150 g is equal to 0.15 kg. since 1000 g = 1kg. </em></u>

<u><em>m = 0.15 kg</em></u>

<u><em></em></u>

<u><em>The dropped from 3.0 meters is irrelevant as the question tells us the initial velocity of the egg: 4.4 m/s before it hits the ground.</em></u>

<u><em>v₀ = 4.4 m/s [down]</em></u>

<u><em></em></u>

<u><em>When it comes to a stop, the egg will have a velocity of 0.</em></u>

<u><em>v = 0 m/s</em></u>

<u><em></em></u>

<u><em>The time it takes for the egg to stop is 0.072 seconds.</em></u>

<u><em>Δt = 0.072 s</em></u>

<u><em></em></u>

<u><em>Therefore, if down is positive, then</em></u>

<u><em></em></u>

<u><em>   </em></u>

<u><em></em></u>

<u><em>We round to two significant figures since every quantity has two sig. figs.</em></u>

<u><em>We only care about the magnitude, not direction. The answer is 9.2 N.</em></u>

<u><em>Unlimited, ad-free access to all of the questions with Brainly Plus</em></u>

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2 years ago
Which describes Einstein’s second postulate about the special theory of relativity? The speed of light in a vacuum is constant,
crimeas [40]

Answer:A

Explanation:

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2 years ago
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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago
Maurice directs a beam of light on a sheet of glass at an angle of 51°. What is the angle of refraction in the glass? The refrac
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2 years ago
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5. You head downstream on a river in an outboard.
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Answer:

hope this helps you're welcome

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2 years ago
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