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ludmilkaskok [199]
3 years ago
14

A ball is rolling across the floor at a constant speed. What will happen to the ball if it is exposed to an unbalanced force in

the same direction that it is moving?
A. It will immediately stop.
B. Its speed will increase.
C. Its speed will decrease.
D. Its motion will be unchanged.
Physics
2 answers:
KATRIN_1 [288]3 years ago
7 0

Answer:

B. it's speed will increase

lions [1.4K]3 years ago
4 0

Explanation:

B. Its speed will increase.

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(a) a fundamental law of motion states that the acceleration of an object is directly proportional to the resultant force exerte
Musya8 [376]

The dimension of force, F is ML/T².

<h3>What is force?</h3>

Force is a push o push agent which causes a change in the state of rest or motion of an object.

From a fundamental law of motion states that the acceleration of an object is directly proportional to the resultant force exerted on the object and inversely proportional to its mass.

Mathematically; acceleration ∝ F/m

F = ma

dimension of Mass = M

dimension of acceleration = L/T²

dimension of force, F = ML/T²

In conclusion, the dimension of force is obtained from the dimensions of mass and acceleration.

Learn more about dimension of force at: brainly.com/question/28243574

#SPJ1

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2 years ago
Where could you hear a sonic boom
ExtremeBDS [4]
Answer

A sonic boom is a continuous effect that occurs while the object is travelling at supersonic speeds.
7 0
4 years ago
Give an example of an energy conversion that produces an enwanted from of energy
stira [4]
An example of an energy conversion that produces an unwanted form of energy is mixing acids with water. 
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3 years ago
A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr
fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}

The values are

\displaystyle y_o=15\ m

\displaystyle v_o=14\ m/s

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

\displaystyle y=0

\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

Multiplying by 2

\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

3 0
3 years ago
PLEASE HELP ME VERY IMPORTANT
saul85 [17]

Answer: let's see, so it's a 1/4 chance of getting it right, hope your odds are good, have a wonderful day

Explanation:

:D

6 0
3 years ago
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