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2 years ago

List 2 examples of objects demonstrating mechanical energy.

1 answer:

6 0

Windmills run on the principle of mechanical energy and work. Moving air (wind) possesses some amount of energy in the form of kinetic energy (due to motion). This energy gives the air the ability to do work on the blades of the fan.

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Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa

34kurt

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

$\mathsf{v^2 = u^2 + 2as}$

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

$\mathsf{v^2 - u^2 =2as }$

$\mathsf{a = \dfrac{v^2 - u^2 }{2s}}$

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

$\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}$

$\mathsf{a = \dfrac{- 771.7284 }{2}}$

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

$\mathsf{t = \dfrac{-u}{a}}$

$\mathsf{t = \dfrac{-27.78}{-385.86}}$

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

brainly.com/question/11298125?referrer=searchResults

3 0

3 years ago

What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N

Varvara68 [4.7K]

We will have the following:

$\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}$

So, the force is approximately 1.85*10^-6 N.

7 0

1 year ago

(b)

m_a_m_a [10]

Answer:

Driving force increases, friction forces increase, the driving force is bigger than friction 12.

Explanation:

4 0

3 years ago

What does it mean to say the mass is conserved during a physical change ?

Olin [163]

Mater doesn't just appear or disappeared. Chemical elements are still there just the connections and how it combines changes.

So what goes into your chemical eqation must still exist after the change.

4 0

3 years ago

Read 2 more answers

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

4 years ago