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egoroff_w [7]
3 years ago
14

A car's bumper is designed to withstand a 6.12 km/h (1.7-m/s) collision with an immovable object without damage to the body of t

he car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.210 m while bringing a 810 kg car to rest from an initial speed of 1.7 m/s.
Physics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

5572.8 N

Explanation:

Applying,

F  = ma.............. Equation 1

Where F = Force, m = mass of the car, a = acceleration.

We can find a by applying,

v² = u²+2as............. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration,  = distance.

From the question,

Given: v = 0 m/s (come to rest), u = 1.7 m/s, s = 0.210 m

Substitute these value into equation 2

0² = 1.7²+2×0.21×a

a = -1.7²/(2×0.21)

a = -2.89/0.42

a = -6.88 m/s²

Also given: m = 810 kg

Substitute these value into equation 1

F = 810(-6.88)

F = -5572.8 N

Hence the force on the bumber is 5572.8 N

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Answer:

6 bits

Explanation:

The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.

Also, Sampling frequently (also known as oversampling) can help in improving signal quality.

To get the number of bits, we use:

2ⁿ = amplitude level

where n is the number of bits.

Given an amplitude level of 64, hence:

2ⁿ = 64

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3 years ago
_____ developed the universal law of gravitation.
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Answer: Newton

Explanation:

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4 years ago
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Jim pushes hard on a rock that weighs 500 pounds. He pushes with a force of 50 pounds, but the rock doesn't move. How much work
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Answer:

Zero work

Explanation:

Recall that work done on  an object is defined as the product of the net force applied to an object times the distance it travels in the same direction as the force.

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3 years ago
A 5kg object is moving downward at a speed of 12m/s. If it is currently 2.6m above the ground, what is its potential energy? Use
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Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c
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Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

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           $=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

         

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