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3 years ago

A golf ball is rolling in the grass. What must happen to stop the ball from continuing to roll?

1 answer:

4 0

The net force must be zero

This is in accordance to Newton's first law, which states that any object in motion will remain in motion and any object at rest will remain at rest unless acted upon by an unbalanced force. An unbalanced force is one where the net force is not zero. If no unbalanced force is applied to a moving object, it will keep moving forever. The reason that we do not observe this in our daily lives is due to friction acting as the unbalanced force.

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The population of oak trees in a forest ecosystem is decreasing. The population of squirrels will most likely _____.

kondor19780726 [428]

The answer should be ''Decrease'' to fill in the blank because squirrels usually live in trees and get food from there so without a place to live when its raining and no food to survive, than they would die and lessen their population so that's why decrease is the answer.

7 0

3 years ago

A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)

lord [1]

Answer:

4.53482 m/s

4.506 m/s

Explanation:

$m_1$ = Mass of player = 75 kg

$v_1$ = Initial velocity of player = 4.6 m/s

$m_2$ = Mass of ball = 0.47 kg

$v_1$ = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

$(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s$

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

$(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s$

The player's speed is 4.506 m/s

4 0

4 years ago

If total (PE + KE) is considered

diamong [38]

The sum of Potential Energy and Kinetic Energy is called Mechanical Energy

3 0

4 years ago

A man travels 8,000 meters in 20 mins how fast is he going

Elden [556K]

Speed = 8000m/20min = 400m/min

Convert meters/min to meters/second.
60seconds in 1minute:
400m/60sec = 6.67m/s

Speed is: 6.67m/s

5 0

3 years ago

Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail

riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of 1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, $\underset{d}{\rightarrow}$ = 10 m + 10 m - 20 m = 0 m

7 0

3 years ago