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Anestetic [448]

3 years ago

12

7) List three (3) automobile safety features currently used to minimize the risk of injury to its passengers. Relate these

1 answer:

r-ruslan [8.4K]3 years ago

3 0

Seatbelt- strapped in egg
air bag- cushion around the egg
brakes- parachute(bag that helps the egg go down slower)

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How much kinetic energy is in a punch thrown at 30 m/s? The fist and arm weighs 6 lbs. (1 lb= 2.2 kg)

zvonat [6]

Answer:

5940J

Explanation:

KE = 1/2 mv²

KE = 1/2 13.2 * 30²

KE = 6.6*900

KE = 5940J

3 0

3 years ago

Objects can be charged by the transfer of electrons.<br> True<br> False

Sveta_85 [38]

Answer:

True

Explanation:

Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. Three ways electrons can be transferred are conduction, friction, and polarization. In each case, the total charge remains the same.

I tried, hope this helps :)

* I might be wrong though

4 0

4 years ago

The average power of a wind turbine is 3 MW.

IgorC [24]

Answer:

a

Explanation:

5 0

3 years ago

Read 2 more answers

High speed stroboscopic photographs show that the head of a 244 g golf club is traveling at 57.6 m/s just before it strikes a 45

almond37 [142]

Answer:

The speed will be "1.06 m/s".

Explanation:

The given values are:

Momentum,

m1 = 244 g

m2 = 45.2 g

On applying momentum conservation ,

Let v2 become the final golf's speed.

From Momentum Conservation

⇒ $Total \ initial \ momentum = Total \ final \ momentum$

⇒ $m1\times u1 + m2\times u2 = m1\times v1 + m2\times v2$

On putting the estimated values, we get

⇒ $0.244\times 57.6+0=0.244\times 39.9+45.2\times v2$

⇒ $57.844+0=9.7356+45.2\times v2$

⇒ $48.1084=45.2\times v2$

⇒ $v2=\frac{48.1084}{45.2}$

⇒ $v2=1.06 \ m/s$

6 0

3 years ago

A 3500 kg spaceship is in a circular orbit 220 km above the surface of Earth. It needs to be moved into a higher orbit of 360 km

Vitek1552 [10]

Answer:

The work done is calculated as $0.044\times 10^{11}\ J$

Solution:

As per the question:

Mass of the spaceship, m = 3500 kg

Height of the orbit, h = 220 km

Mass of the earth, $M_{e} = 5.97\times 10^{24}\ kg$

Height of the higher orbit, h' = 360 km

Radius of the orbit, $R = 6.37\times 10^{6}\ km$

Now,

The work done is given by the change in the gravitational potential energy:

Gravitational Potential Energy, U = $- \frac{GM_{e}m}{R + h}$

Now, for the lower orbit:

$U_{l} = - \frac{GM_{e}m}{R + h}$

$U_{l} = - \frac{6.67\times 10^{- 11}\times 5.97\times 10^{24}\times 3500}{6.37\times 10^{6} + 220\times 10^{3}}$

$U_{l} = - 2.115\times 10^{11}\ J$

For upper orbit:

$U_{U} = - \frac{GM_{e}m}{R + h}$

$U_{u} = - \frac{6.67\times 10^{- 11}\times 5.97\times 10^{24}\times 3500}{6.37\times 10^{6} + 360\times 10^{3}}$

$U_{U} = - 2.071\times 10^{11}\ J$

Change in the gravitational Potential energy:

$\Delta U = U_{U} - U_{l} = - 2.071\times 10^{11} - (- 2.115\times 10^{11}) = 0.044\times 10^{11}\ J$

Therefore, the work done:

$W = 0.044\times 10^{11}\ J$

7 0

3 years ago