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Inessa05 [86]
3 years ago
15

A student determines a value for a force using the formula F = k I l.

Physics
1 answer:
kramer3 years ago
5 0

Answer:

maybe dyne.......................

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A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
Naddika [18.5K]

The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

F = \frac{mv}{t}

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

F = \frac{0.07 \times 56}{0.04}

F = \frac{3.92}{0.04}

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: brainly.com/question/13590154

4 0
2 years ago
Read 2 more answers
Please help me fill out the chart.??
LekaFEV [45]

Answer:unbalanced: have direction,Change an objects motion, causes object to accelerate

Balanced:Do not change an objects motion, Net forces equal sum of all forces on object, and Does not equal 0 N

Explanation: Thats all I know Hoped I helped sum

5 0
3 years ago
WILL MARK BRAINLIEST!
boyakko [2]
It’s mass is 100 kg, because if you divide both numbers, you get 100
6 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
What force must be used to do 224 Joules of work on an object over a distance of 32 meters?
cestrela7 [59]
7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons
4 0
3 years ago
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