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2 years ago

How could a slow moving train had the same momentum as a high-speed bullet

Physics

1 answer:

Flura [38]2 years ago

6 0

Answer:

it is sooo easy u need to use magnetic panels on the rail and put the magnet on the train it works under the principal of magnetic

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If a ball is dropped from a height (H) its velocity will increase until it hits the ground (assuming that aerodynamic drag due

mario62 [17]

Answer:

9.801 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 39 ft/s

s = Displacement = 720 cm = 7.2 m

a = Acceleration

Converting to m/s

$39\ ft/s=\frac{39}{3.281}=11.88\ m/s$

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{11.88^2-0^2}{2\times 7.2}\\\Rightarrow a=9.801\ m/s^2$

Acceleration of the ball is 9.801 m/s²

5 0

3 years ago

How to calculate the unknown force

Lesechka [4]

the formula to calculate the force is, F = MA

5 0

2 years ago

When a unbalanced force act on an object , the object moves with what motion?

jeka94

There's no such thing as "an unbalanced force".
When the entire group of forces acting on an object is unbalanced,
meaning that the net force on the object is not zero, then the object
moves with accelerated motion . . . it may speed up, it may slow down,
or its speed may remain constant while its direction changes.

3 0

3 years ago

A bond between a positively-charged particle and a negatively-charged particle is a(n) _______ bond.

weeeeeb [17]

The answer is an ionic bond.

6 0

3 years ago

People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p

Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

$g' = \omega^2 (\dfrac{r}{2})$

$g' =\dfrac{1}{2}(\omega^2 r)$

$g'=\dfrac{g}{2}$

People at the halfway experience g/2

hence, the correct answer is option B

7 0

3 years ago