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1 year ago

A student drops a ball from a slope. He notices the ball rolling down as shown in the image.

Physics

1 answer:

zavuch27 [327]1 year ago

5 0

Answer:

You are correct, it is (3) the ball is in motion as it rolls down the slope.

Explanation:

For number 1 it would make sense because how would the student be in motion if he is just standing there throwing the ball down the slope, not himself.

For number 2 still wouldn't make so much sense since we are talking about the ball being in motion not the student.

For number 3 it seems the most reasonable because the ball is in motion as it is rolling down the slope.

For number 4 it doesn't make sense for a ball to lose its shape just by rolling, also even if you're like: "oh, but what if its leaking air or has a hole??" well easy then. It STILL wouldn't matter because in the sentence it never specifies it has a "hole" or is "leaking air" therefore this answer is wrong.

YOUR ANSWER IS number 3.

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Which of the following is the best description of a gaseous substance?

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Answer:

It has no shape of its own but has a definite volume.

Explanation:

Gases have no shape but a definite volume

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alexdok [17]

Answer:

$30 N \cdot m$

Explanation:

The torque applied by a force can be calculated as

$\tau = F d sin \theta$

where

F is the magnitude of the force

d is the length of the arm

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

In this problem, we have

F = 15 N

d = 2.0 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m$

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2 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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2 years ago

A carpenter lifts a 10-kg piece of wood to his shoulder 1.5 m above the ground. He then sets the wood on his truck at 1.0 m abov

FinnZ [79.3K]

Answer:

500Joules

Explanation:

Kinetic energy = 1/2mv²

m is the mass of the wood

v is the velocity

Given

Mass = 10kg

Velocity v = 10m/s

Substitute into the formula and get KE

KE = 1/2 * 10 * 10²

KE = 1/2 * 1000

KE = 500Joules

Hence the kinetic energy of the wood during delivery is 500Joules

8 0

2 years ago

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b

lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

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radius(r)=2.5 m

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Here Frictional force will balance weight

And limiting frictional force is provided by Centripetal force

$f=\mu N=\mu m\omega ^2\cdot r$

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Equating two

f=mg

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$\omega ^2=\frac{g}{\mu r}$

$\omega =\sqrt{\frac{g}{\mu r}}$

$\omega =2.55 rad/sec$

8 0

2 years ago