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2 years ago

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

Physics

1 answer:

Sergeu [11.5K]2 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 small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0

GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

$E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

$F=qE$

Put the value into the formula

$F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

We need to calculate the total charge on the rod

$Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}$

Put the value into the formula

$Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}$

$Q=47.8\times10^{-9}\ C$

$Q=47.8\ nC$

Hence, The total charge on the rod is 47.8 nC.

6 0

2 years ago

Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They

denis-greek [22]

Answer:

v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

Em₀ = U = m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

m g h = ½ m v²

v² = 2gh

v = √ 2gh

Let's calculate

v = √ (2 9.8 150)

v = 54.2 m / s

3 0

2 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

2 years ago

If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over

astra-53 [7]

Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the friction force keeping on table is given by the following formula:
Ff = u*(1-f)*m*g 
and we need to know aso that gravity force pulling off the table Fg is given by this other formula:
Fg = f*m*g 
What you need to do is Equate the two and solve for f: 

f*m*g = u*(1-f)*m*g 
=> f = u*(1-f) = u - uf 
=> f + uf = u 
=> f = u/(1+u) = fraction of rope
With that you can find the answer

8 0

2 years ago

Read 2 more answers

Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic

Alinara [238K]

The volume of the room is the product of its dimensions:

$10\times 12 \times 4 = 480\text{ m}^3$