A bowling ball with a mass of 7.10 kg has a velocity of 5 m/s. what is the bowling ball's momentum?

A pure copper sphere has a radius of 0.935 in. How many copper atoms does it contain? [The volume of a sphere is (4>3)pr3 and

ValentinkaMS [17]

Answer:

47.68 x 10²³ or 4.768 x 10²⁴ copper atoms

Explanation:

Given:

Radius of the copper sphere (r) = 0.935 in

First convert the radius from inches to centimeters

1 in = 2.54cm

0.935in = 0.935 x 2.54cm = 2.3749cm

∴ r = 2.3749cm

Calculate the volume of the copper sphere as follows

Volume = $\frac{4}{3}\pi r^{3}$ [substitute r = 2.3749cm and π = 22 / 7]

Volume = $\frac{4}{3}(\frac{22}{7} ) (2.3749)^{3}$

Volume = 56.108cm³

From the volume and given density, calculate the mass of the copper sphere

mass = density x volume [density = 8.96g/cm³]

mass = 8.96 x 56.108 = 502.73g

From known facts

1 mole of copper = 63.5g of copper = 6.022 x 10²³ copper atoms.

Then,

502.73 g of copper = $\frac{502.73 * 6.022*10^{23}}{63.5}$ = 47.68 x 10²³ copper atoms

Therefore, the sphere contains 47.68 x 10²³ copper atoms

6 0

4 years ago

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A man pulls on a (massless) rope tied to a tree with a force of 500 N. Later, two men pull on opposite ends of the same rope wit

qwelly [4]

Answer:

In both cases the tension in the rope is equal to 500N

Explanation:

It may be that in the case of the tree, the result is more intuitive, because you can think that there is only one force. But this is misleading.

To find the tension in the rope, you should draw a free body diagram. By doing so, you would find that the rope is static because there are two opposite forces. Assuming, for simplicity, that the rope is horizontal, a force of 500N is pulling to one direction (let's say to the right) and a force of 500N is pulling to the opposite direction (to the left). Else, the rope would not be static.

That analysys is the same for the rope tied to the tree ( the tree is pulling with 500N, such as the man, but in opposite direction) and when the rope is pulled by two men on opposite ends, each with forces of 500N.

Hence, the tension is the same and equal to 500N.

7 0

3 years ago

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A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$

F) -11.1 m/s

The y-component of the velocity at time t is given by:

$v_y(t) = v_y + at$

where

$v_y = 9.3 m/s$ is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

$v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s$

where the negative sign means the vertical velocity is now downward.

3 0

3 years ago

The force of gravity between two objects depends on the objects’ mass and (4 points) Select one: a. their shapes b. what they ar

konstantin123 [22]

There is an equation for this, and it will help us answer the question:

F = G × m₁*m₂ ÷ r²

The question already tells us the force of gravity between two objects depends on the objects' mass. Now we have to find a second factor.

Their shapes are not a secondary factor to this question.

What they are made of is not a factor to the gravitational force between the two.

Their weights have nothing to do with this either.

However, the distance does. According to the equation

F = G × m₁*m₂ ÷ r²

r² is the radius, or in other words, the distance between the two masses from their center of gravity. Therefore the distance and masses of two objects are what affect the gravitational pull between the two.

Have a nice day!

5 0

4 years ago

If you have a density of 5 g/mL and a mass of 2.5 g, what is the volume?

Korvikt [17]

Density = Mass/Volume
2.5 g/ml = 5g/Volume
Volume = 5g/(2.5 g/ml)
= 2 ml

5 0

3 years ago