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KonstantinChe [14]
4 years ago
15

A bow accelerates a 100g arrow at 50m/s calculate the force acting on the arrow in Newton’s

Physics
1 answer:
Norma-Jean [14]4 years ago
5 0

Answer:

Explanation:

Newton's law states that F=mg, or in other applications F=ma, where:

F= force acting on the system  

m= mass of the system and

a= acceleration (or g= gravitational acceleration which is 9.8m/s^2)

In your question we know the following variables:

The mass of the arrow (m= 100g, however we should use kilograms- 0.1kg)

The acceleration of the arrow (a=50m/s^2)

If we substitute these values into our given equation, we can find the force acting on the arrow.

F = m x a

F= (0.1) x (50)

F = 5N

hope this helps :)

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Which of the following is a way that some animals bodies adapt to living in extremely cold climates A. Clusters B. Hibernation C
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3 0
4 years ago
A pulley in the shape of a solid cylinder of mass 1.50 kg and radius 0.240 m is free to rotate around a horizontal shaft along t
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Answer:

the speed of the textbook just before it hits the floor is 2.4 m/s

Explanation:

  Given the data in the question;

mass of pulley = 1.50 kg

radius of pulley = 0.240 m

mass of text book = 2.0 kg

height from which text book was released = 0.9 m

angular speed of the pulley = 10.0 rad/s

the speed of the textbook just before it hits the floor = ?

the speed of the textbook v = angular speed of the pulley × radius of pulley

we substitute

v = 10.0 rad/s × 0.240 m

v = 2.4 m/s

Therefore, the speed of the textbook just before it hits the floor is 2.4 m/s

6 0
3 years ago
A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a se
patriot [66]
<h2>Answer:20.97g N,32.63g N</h2>

Explanation:

We consider the forces at the knot.

The vertical forces are

T_{2}Sin(50^{0}}) is the vertical component of tension T_{2} at the knot.

-25g is the weight of the mass 25Kg acting downwards.

The horizontal forces are

-T_{1} is the tension in the rope acting left.

T_{2}Cos(50^{0}) is the horizontal component of tension T_{2} acting towards right.

Since the knot has no mass,it is always in equilibrium.

So,the sum of forces acting on it will be zero.

Balancing vertical forces gives,

T_{2}Sin(50^{0}})-25g=0

T_{2}=32.63gN

Balancing horizontal forces gives,

T_{2}Cos(50^{0})-T_{1}=0

T_{1}=32.63g\times Cos(50^{0})=20.97gN

7 0
3 years ago
Read 2 more answers
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