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3 years ago

Explain why it is dangerous to jump from a fast moving train

Physics

1 answer:

Sergeu [11.5K]3 years ago

7 0

Answer:

When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.

Explanation:

you be in serious injury.

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A 480N sphere 40.0cm in radius rolls without slipping 1200cm down a ramp that is inclined at 53 0 with the horizontal. What is t

IgorC [24]

As we know that sphere roll without slipping so there is no loss of energy in this case

so here we can say that total energy is conserved

Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'$

as we know that ball start from rest

$v_i = 0$

height of the ball initially is given as

$h = Lsin\theta$

$h = 1200sin53 = 960 cm$

also we know that

$I = \frac{2}{5}mR^2$

also for pure rolling

$v = r\omega$

also we know that

$480 = m*9.8$

$m = 49 kg$

now plug in all data in above equation

$480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0$

$4608 = 3.92\omega^2 + 1.568\omega^2$

$\omega^2 = 839.65$

$\omega = 29 rad/s$

So speed at the bottom of the inclined plane will be 29 rad/s

7 0

3 years ago

Speedy, the super-swift squid, swims sideways at 5.3 m/s. If he suddenly uses his ink jet to accelerate at 15 m/s/s for 2.1 s, w

san4es73 [151]

Use v = vi + at
v = 5.3 m/s + (15 m/s^2)(2.1 s)
v = 5.3 m/s + 31.5 m/s
v = 36.8 m/s

7 0

3 years ago

A 5.0 mm diameter proton beam carries a total current of 1.5mA. The current density in the proton beam, which increases with dis

slavikrds [6]

Answer:

Jedge = 0.076A/cm^2

Explanation:

If you want to know what is Jedge is convenient to use an integral. The total density current is given by:

$J_T=\int_{0}^{R}J_{edge}(\frac{r}{R})dr=J_{edge}(\frac{1}{R})(\frac{R^2}{2})=\frac{J_{edge}R}{2}$ (1)

But also, we have that the total current is

$I_T=J_TA\\\\J_T=\frac{I_T}{A}=\frac{1.5*10^{-3}A}{\pi(5*10^{-3})^2}=19.09\frac{A}{m^2}$

where we have used that 5mm=5*10^{-3}m.

By replacing (1) we obtain:

$J_{edge}=\frac{2J_T}{R}=\frac{2(19.09A/m^2)}{5*10^{-3}m}=7639.4\frac{A}{m^2}=0.076\frac{A}{cm^2}$

hope this helps!!

3 0

4 years ago

Read 2 more answers

Please Help!<br><br><br>What is inertia? What does inertia depend on?

Lemur [1.5K]

Inertia is that quantity which depends solely upon mass. The more mass, the more inertia. Momentum is another quantity in Physics which depends on both mass and speed.

8 0

2 years ago

The tip of a nail exerts tremendous pressure when hit by a hammer because it exerts a large force over a small area. What force

yanalaym [24]

Answer:

Force, F = 2714.33 N

Explanation:

It is given that,

Diameter of the circular tip, d = 1.2 mm = 0.0012 m

Radius, r = 0.0006 m

Pressure acting on the hammer, $P=2.4\times 10^9\ N/m^2$