An electric lamp transfers 500j in 5s what is it’s power

An empty train car of mass 2.0 x 10^4 kg coasts along at 10 m/s. A 3000-kg boulder is suddenly dropped vertically into the car.

Zepler [3.9K]

Answer:8.69 m/s

Explanation:

Given

mass of Empty train $m=2\times 10^4 kg$

velocity of car $v=10 m/s$

mass of Boulder $M=3000 kg$

Since no External force is acting therefore conserving Momentum

Train and boulder moves with same velocity after collision

$m v=(M+m)v'$

$2\times 10^4\times 10=(2\times 10^4+3000)v'$

$v'=\frac{20,000\times 10}{23000}$

$v'=8.69 m/s$

8 0

3 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0

3 years ago

Explain how if a solid, liquid, and gas are put into individual containers how they fill the container.

Sav [38]

Answer:

serial in which container is filled

Solid -base of container

Liquid- above solid

Gas- above liquid

Explanation:

If any mixture of matter in different state (that solid , liquid or gas )are kept in any container, then substance with higher density will be settled at lowest surface first and similarly the substance with lowest density will be at upper part of container.

In the given container we have to keep solid, liquid and gas

sold has the highest density,
gas the lowest density and
liquid has the density higher than gas but less than solid.

based on this

solid will be at surface of container

above sold will be liquid

above liquid will be presence of Gas

serial in which container is filled

Solid -base of container

Liquid- above solid

Gas- above liquid

8 0

4 years ago

What does your body do with nutrients during the digestion process?

kkurt [141]

Answer:

the correct answer is

Explanation:

The small intestine absorbs most of the nutrients in your food,and your circulatory system passes them on to other parts of your body to store or use. Special cells helped absorbed nutrients cross the intestinal lining into your blood stream.

hope this works out!!!!

6 0

3 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

4 years ago