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3 years ago

What is the SI unit for graviational potential energy

Physics

1 answer:

scZoUnD [109]3 years ago

3 0

Gravitational potential energy is energy.
The unit of energy is the Joule.

1 Joule = 1 kilogram-meter² / sec²

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A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s.

Katarina [22]

Option B would be right one
according to momentum conservation
6600*2 = 13200kgm/s
5400*3 = 16200kgm/s
16200-13200 = 3000
now 6600-5400 = 1200 kg
thus 3000/1200 = 2.5 v

5 0

3 years ago

Which of the following items best embodies the physical property of conductivity?

Ostrovityanka [42]

Copper penny is the answer

8 0

3 years ago

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A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0

3 years ago

An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =

fiasKO [112]

Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

Explanation:

To find the coordinates of event in system K ,we have to use inverse Lorentz transformation

$x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m$

for t

$r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2} )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s$

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

6 0

3 years ago

A student pushes on a 5 kg box with a force of 20 N forward. The force of sliding friction

mart [117]

Answer:

Explanation:

F=20N

mu=-10N

F=ma

20-10=10

10=(5)(a)

a=2 m/s^2

4 0

3 years ago