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klio [65]
2 years ago
13

A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall

of the tool compartment. If the wrench stops after being in contact with the wall for 0.07 s, what is the average force exerted on the wrench by the wall
Physics
1 answer:
Naddik [55]2 years ago
8 0

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

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3 years ago
Can somebody help me please
Eddi Din [679]
51 inches.

This is because a stem-plot is formatted as so:

If it’s 5 on the left side of the line, anything on the right is the ones place making possible numbers 51, 53, 56, etc.

Hope this helps!
3 0
3 years ago
The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of
Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is  =4.51m/s  

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

        Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

  Now to obtain the initial energy stored

               Let U denote the initial energy stored and

                      U  = \frac{1}{2} kx^2

Where x  is the length the spring is displaced

k is the force constant of the string

         U = \frac{1}{2} * 627 * (0.6)^2

          = 112.86 J

   Now referring to the formula above

i.e          Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

                \frac{1}{2} mv^2 = 112.86 - \mu_kmgx

                v^2 = \frac{2(112.86 -\mu_kmgx)}{m}

                v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}

and we are told that coefficient of friction  = 0.4 and the mass is 9 kg ,the acceleration due to gravity = 9.8m/s^2  this displacement length of spring = 0.6

  Therefore   v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }

                        =4.51m/s      

           

8 0
3 years ago
1 What conclusion can you draw about semicircular canals from their name ? A Their shape B Their size Their location in your bod
oee [108]

Answer:

A. Their shape

Explanation:

The name clearly shows that the shape of these canals are semi-circular. The semi-circular canals consist of three tubes filled with fluid and located in the inner ear. They help to maintain balance and transmit impulses through the movement of the fluids. The impulses sent through these fluids are sent to the brain for interpretation.

The function of the canals are not indicated by the name, rather the shape is hinted through the name.

7 0
2 years ago
Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th
DENIUS [597]

Answer:

a) vboat = 5.95 m/s  b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As  we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w =  \frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
3 years ago
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