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3 years ago

Why is it important for engineers to design bumper cars safely

2 answers:

7 0

Answer:if they don’t design the bumper cars safely then people could get hurt while using them because they could break.

Explanation:

rosijanka [135]3 years ago

3 0

Answer:

Cars have bumpers designed to protect the body of the car from minor damage during low-speed collisions. ... They will use the engineering design process to design and build bumpers to protect the main parts of their car from damage, and use their knowledge of Newton's third law to explain what they observe.

Explanation:

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Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a

dusya [7]

Answer:

$W=-21,870,000\ J$

Explanation:

Work and Kinetic Energy

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

$W=\Delta k=k_1-k_0$

Since

$\displaystyle k=\frac{mv^2}{2}$

We have

$\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}$

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

$\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}$

$W=0-21870000\ J$

$\boxed{W=-21,870,000\ J}$

3 0

3 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

Which part of a wind-powered system ultimately produces the electricity? A. nacelle B. blade C. turbine D. generator

djverab [1.8K]

The correct answer is: (D) Generator

Explanation:
In wind-powered systems, the wind energy turns the blades around the rotor of a wind turbine. That rotor is connected to a generator that generates electricity. In other words, the kinectic energy of the wind is converted into electrical energy by using the generator in the wind-powered systems.

5 0

2 years ago

Read 2 more answers

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m

denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0

2 years ago

A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

Morgarella [4.7K]

Answer:

B.0

Explanation:

Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s

Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

Change in momentum = m(v-v)

Change in momentum = m(0)

Change in momentum = 0 kg.m/s

Therefore the momentum of the ball has not changed.

The right option is B.0

5 0

3 years ago