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Arturiano [62]
3 years ago
7

At the instant that the velocity of the crate is v⃗ =(3.40m/s)ι^+(2.20m/s)j^, what is the instantaneous power supplied by this f

orce?
Physics
1 answer:
Zigmanuir [339]3 years ago
5 0
I found some missing information about this problem online. We are given the force:
F = F =(-7.50N)i +(3.00N)j
Power is defined as the rate of doing work. 
This is the formula:
P= \frac{dW}{dt}
Where P is power, W is work. 
Work is defined as:
W=F\cdot r
F is the force and r is the displacement.
If we assume that force is not changing (it's constant) with time we get:
P= \frac{dW}{dt}=F \frac{dr}{dt}=F\cdot v
Keep in mind that both force and velocity are vectors, so we have to multiply each component separately.
Finally, we get:
P=F_i\cdot v_i+F_j\cdot v_j=(-7.50N)(3.40\frac{m}{s})+(3.00N)(2.20\frac{m}{s})=
-18.9 W


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Explanation:

From the question we are told that

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  The mass  of the wood is  m_w =  900 \  g  =  0.90\  kg

   The height attained by the combined mass is  h =  8.0 \ cm  =  0.08 \ m

Generally according to the law of energy conservation    

    KE _b  =  PE_c

Here KE_b is the kinetic energy of the bullet before collision.

and PE_c is the  potential  energy of the combined mass of bullet and wood at the height h which is mathematically represented as

      PE_m  =  [m_b  + m_w] *  g *  h

So

   KE_b =PE_c   = [0.005  + 0.90] * 9.8 *0.08

=> KE_b =0.710 \ J

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3 years ago
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3 years ago
What is the magnitude of the kinetic frictional force
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4 0
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a
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The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

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3 years ago
Read 2 more answers
Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh
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Answer:u=\frac{v}{2}\sqrt{5-4sin\phi }

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and solving problem in Vertical and horizontal direction

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4u^2=5v^2-4v^2sin\phi

u=\frac{v}{2}\sqrt{5-4sin\phi }

7 0
3 years ago
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