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2 years ago

2.4 physics assignment.

Physics

1 answer:

kvv77 [185]2 years ago

4 0

Answer:The distance o the ramp that the car traveled is given by d=(1/2)at^2=(0.5)(3.96)(5.76)^2=65.69 meters. The horizontal component of this travel is 65.69*

Explanation:

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CHALLENGE

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When the tractor pulls, the boy moves. When the elephant moves, the boy moves. If the elephant moves and has more power than the tractor, the tractor and the boy moves. If the tractor moves and has more power than the elephant, then the elephant and the boy moves. I hope this helps you, have a great day!

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3 years ago

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Maceo is making rock candy. Which best describes the steps she should take?

Andru [333]

Answer:

The answer is heat a saturated sugar water solution, dissolve more sugar, then let the solution cool

Explanation:

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3 years ago

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The atmosphere of Mercury and Mars are very thin. What effect does the thin atmosphere have on the temperature on the surface of

KengaRu [80]

Answer:

Very hot during the day and very cold at night.

Explanation:

Due to the thin atmosphere, they have very hot climate during the day time and very cold climate at night. This happens because they contain very low amounts of greenhouse gases. These gases retain the heat at night. The atmosphere also prevents excessive light and UV rays from entering. The thin atmosphere leads to many asteroids and comets hitting the surface of the planet. On earth, these asteroids usually, burn up in the mesosphere layer of the atmosphere. These asteroid collisions cause massive fires. This in turn, causes the temperature to increase during the day. During the night time, massive fires cannot burn due to the low temperature because of the lack of greenhouse gases.

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3 years ago

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

$F = \frac{KQ_{1}Q_{2}}{d^{2}}$

$1.60 = \frac{4KQ^{2}}{d^{2}}$ .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

$F' = \frac{9KQ_{3}Q_{4}}{d^{2}}$

$F' = \frac{126KQ^{2}}{d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0

3 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

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3 years ago