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bogdanovich [222]
3 years ago
14

A 0.45-kilogram football traveling at a speed of 22 meters per second is caught by an 84-kilogram stationary receiver. if the fo

otball comes to rest in the receiver's arms, the magnitude of the impulse imparted to the receiver by the ball is
Physics
1 answer:
Troyanec [42]3 years ago
3 0

Impulse = Change in momentum.  
The ball was moving with a momentum of 0.45 * 22 = 9.9  
The ball comes to rest in the receivers arm; this means the ball's final velocity = 0. So mv2 = 0.45 * 0  
The magnitude of the impact is just the change in momentum. 9.9 - (0.45 * 0) = 9.9
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In a lever, the effort arm is two times as long as the load arm. The resultant force will be
jasenka [17]
Answer is B. 

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

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Charlie
5 0
3 years ago
Read 2 more answers
In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:
vesna_86 [32]

Answer:

D. 1×10⁻⁸ M

Explanation:

[H⁺] [OH⁻] = 10⁻¹⁴

(1×10⁻⁶) [OH⁻] = 10⁻¹⁴

[OH⁻] = 1×10⁻⁸

6 0
3 years ago
A 7 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The ki
Marizza181 [45]

Answer:

Explanation:

Given

mass of object m=7 kg

kinetic Energy k=36\ J

Tension in string T=326\ N

mass is moving in a horizontal circle so tension is providing the centripetal acceleration

therefore T=\frac{mv^2}{r}----1

where r=radius of circle

kinetic energy of particle k=\frac{1}{2}mv^2----2

divide 1 and 2 we get

\frac{T}{k}=\frac{\frac{mv^2}{r}}{\frac{1}{2}mv^2}

\frac{T}{k}=\frac{2}{r}

r=\frac{2k}{T}=\frac{2\times 36}{326}

r=0.2208\ m

   

8 0
3 years ago
Material
elixir [45]

Answer:

write your question properly

5 0
3 years ago
Sometimes, I have to take my cat, Gigi, in the car. She doesn’t like this, and tends to hide under the couch when she knows she
Reika [66]

Answer:

F - fr = ma ,    N - W = 0

Explanation:

In this exercise we are asked to identify the forces that act on the jack, for this we will use Newton's second law

On the x axis

We have two forces: the friction and the force of the girl who pulls the cat

           F - fr = ma

On the y axis

There are two forces: normal and weight

         N - W = 0

A diagram of these forces can be seen in the attachment

8 0
3 years ago
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