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Oksi-84 [34.3K]
2 years ago
12

It is safer for a car driver to be wearing a seat belt, compared with not wearing a seat belt, if the car is involved in a colli

sion.
Explain why.​
Physics
1 answer:
Korolek [52]2 years ago
8 0

Explanation: It is because when a car is moving both the car and the driver is in inertia of motion. When a car is involved in collision it comes to a sudden stop and the car comes into inertia of rest whereas the person still in inertia of motion moves forward and might result in major injuries. But this can be prevented by wearing a seatbelt

Hope it helps :)

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Two astronauts in space with a baseball decide to play catch to pass the time. In the language of conservation of momentum, desc
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As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
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Explanation:

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The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?
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It takes twice the force to produce the same acceleration in the 2kg rock. 
5 0
3 years ago
How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

            C = \frac{\epsilon_{o}A}{d}

It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

                   = \epsilon_{o}AE

                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
3 years ago
A car of mass 1000 kg travelling at a velocity of 25 m/s collides with another car of mass 1500kg which is at rest. The two cars
Svetach [21]

Answer:

<em>The velocity of the two cars is 10 m/s after the collision.</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and velocity v is

P=m.v

If we have a system of bodies, then the total momentum is the sum of them all

P=m_1v_1+m_2v_2+...+m_nv_n

If some collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum  takes the form:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The car of mass m1=1000 Kg travels at v1=25 m/s and collides with another car of m2=1500 Kg which is at rest (v2=0).

Knowing both cars stick and move together after the collision, their velocity is found solving for v':

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

\displaystyle v'=\frac{1000*25+1500*0}{1000+1500}

\displaystyle v'=\frac{25000}{2500}

v' = 10 m/s

The velocity of the two cars is 10 m/s after the collision.

4 0
3 years ago
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