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Anvisha [2.4K]
3 years ago
15

If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (

3a1)
Question 10 options:

There cannot be any forces applied to the ball.


There must be exactly one force applied to the ball.


The net force applied to the ball is zero.


The net force applied to the ball is directed to the right.
Physics
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

C. The net force applied to the ball is zero.

Explanation:

From Newton's second law of motion;

F = ma

Where F is the force on an object, m is its mass and a is its acceleration.

Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.

Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.

So that;

F = m x 0

  = 0

No force is applied on the object.

Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.

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Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
4 years ago
For a given amount of gas at a constant temperature, the volume of a gas varies inversely with its pressure is a statement of __
Ksivusya [100]

Answer:

d. Boyle's

Explanation:

Boyle's Law: States that the volume of a fixed mass of gas is inversely proportional proportional to its pressure, provided temperature remains constant.

Stating this mathematically. this implies that:

V∝1/P

V = k/P, Where k is the constant of proportionality

PV = k

P₁V₁ = P₂V₂

Where P₁ and P₂ are the initial and final pressure respectively, V₁ and V₂ are the the initial and final volume respectively.

Hence the right option is d. Boyle's

8 0
3 years ago
Which is a likely use for a base? a. as a vitamin in your food b. etching metals for printing c. making foods taste sour d. maki
natta225 [31]
I think the correct answer would be making soaps and detergents. It is the manufacture of soaps and detergents that makes use of a base. It is included in the process of such product. Hope this answers the question. Have a nice day.
5 0
3 years ago
A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?
poizon [28]

Answer:

\boxed {\boxed {\sf 7.84 \ Joules}}

Explanation:

The formula for potential energy is:

PE=m*g*h

where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m

Substitute the values into the formula.

PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)

Multiply the first two numbers.

  • 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
  • If we convert the units now, the problem will be much easier later on.
  • 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

PE=(3.92 \ N )(2 \ m)

Multiply.

  • 3.92 N* 2 m=7.84 N*m
  • 1 Newton meter is equal to 1 Joule (this is why we converted the units).
  • Our answer is equal to<u> 7.84 Joules.</u>

PE=7.84  \ J

6 0
3 years ago
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A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu
satela [25.4K]
<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>
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4 years ago
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