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2 years ago

Topic : Collisions and Conservation of Momentum

1 answer:

7 0

A 0.2 kg ball moves to the right with a speed of 3 m/s. It hits a 0.5 kg ball that at rest. After the collision, the second ball moves to the right with ...

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Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):

$F_{2}=G\frac{(2M)(6M)}{r^2}$ (4)

$F_{2}=G\frac{12M^2}{r^2}$ (5)

$F_{2}=12\frac{GM^2}{r^2}$ (6)

Now, if we isolate $\frac{GM^2}{r^2}$ from (3):

$\frac{F_{1}}{16}=\frac{GM^2}{r^2}$ (7)

Substituting $\frac{GM^2}{r^2}$ found in (7) in (6):

$F_{2}=12(\frac{F_{1}}{16})$ (8)

$F_{2}=\frac{12}{16}F_{1}$ (9)

Simplifying, we finally get the expression for $F_{2}$ in terms of $F_{1}$ :

$F_{2}=\frac{3}{4}F_{1}$

5 0

3 years ago

Which of these is an example of force?

Roman55 [17]

Answer:

B is the answer a force is a push or pull

8 0

3 years ago

When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of

pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0

3 years ago

A bar of gold has a temperature of 31°C, and a bar of aluminum has a

Katarina [22]

Different dense matters

3 0

3 years ago

Calculate the volume of this regular solid.

Nataly [62]

Answer:

V = 42.41cm^3

Explanation:

In order to calculate the volume of the solid, you use the following formula:

$V=\frac{1}{3}\pi r^2 h$

where

r: radius of the circular base of the cone = 3 cm

h: height from the circular base to the peak of the cone = 4.5 cm

You replace the values of r and h in the formula for the volume V:

$V=\frac{1}{3}\pi(3cm)^2(4.5)=42.411cm^3\approx42.41cm^3$

hence, the volume of the solid is 42.41 cm^3

5 0

3 years ago