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dedylja [7]
4 years ago
15

Which of the following is used to determine if a star is moving closer to Earth?

Physics
1 answer:
sergey [27]4 years ago
6 0
This is Doppler Shift 
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What is the kinetic energy of a 14kg object traveling at 10m/s
Sauron [17]

Answer:

Explanation:

kinetic energy=1/2*mass*velocity^2

=1/2*14kg*10^2

=7*100

=700 joule

5 0
3 years ago
- A certain mass is suspended at one end of a spring of spring-constant of 32,000gm/s2. The
babunello [35]

Answer: I am pretty sure it is (b) what is the value of mass suspended at the end of the spring.

Explanation:

6 0
3 years ago
Is an object speeding up or slowing down of the V final is greater than the V initial?
sertanlavr [38]

Answer:

V is greater

Explanation:

because v intial at that time V final is the that speed which it is going at that time

7 0
3 years ago
one mole of water is equivalent to 18 grams of water. a glass of water has a mass of 200 g. how many moles of water is in this?
ehidna [41]
1 mole = 18 g
200 g = glass of water
200 ÷ 18 = 11.1
11.1 moles of water in 200 g (glass of water)
3 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
3 years ago
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