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2 years ago

Why does a passenger jumping out of a rapidly moving bus fall forqard with his face downwards

Physics

1 answer:

Alex Ar [27]2 years ago

6 0

Answer:

Explanation:

It's Newton's first law of motion, and object in motion continues to stay in motion unless acted upon by an unbalanced force. In this case, your body is the force that's moving on the bus. When you jump(the unbalanced force) your body is still used to being in motion, which makes you fall and not land on your feet.

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

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2 years ago

A message is sent from the Galileo spacecraft orbiting Jupiter to earth at a distance of 928,000,000km. If it took the signal 51

expeople1 [14]

The answer would approximately be 299,741.60

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eduard

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172,000 second 

I'M NOT SURE THAT THIS IS RIGHT OR WRONG IF IT'S WRONG THEN SORRY 

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2 years ago

A bag of marbles is hanging motionless on a spring scale. What prevents the bag from falling?

Tanya [424]

Answer:

The vector sum of all forces acting on it is zero, its at equilibrium.

Explanation:

The bag of marbles hanging on a spring scale applies its weight downwards, which was counterbalanced by the reaction from the spring scale (obeying the Newton's third law of motion). And since no external forces are applied to the system, thus the equilibrium of the system.

If the weight of the bag is greater than the reaction from the spring scale, the scale breaks and the system would not be balanced.

7 0

2 years ago

Why does a passenger jumping out of a rapidly moving bus fall forqard with his face downwards​

Why does a passenger jumping out of a rapidly moving bus fall forqard with his face downwards