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maxonik [38]
2 years ago
10

Scientists think the is a solid layer of iron with a liquid layer of iron around it

Physics
1 answer:
statuscvo [17]2 years ago
8 0
I think its true, I think.
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You are traveling in a spaceship at a speed of 0.70c away from Earth. You send a laser beam toward the Earth traveling at veloci
Dovator [93]
The question is asking to describe and state and calculate what do the observer on the earth measure for the speed of the laser beam, and base on my research, the answer would be v = 1bc, I hope you are satisfied with my answer and feel free to ask for more 
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2 years ago
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What is occurring when a light wave goes through a pane of glass in a window?
ycow [4]

Answer:

transmission: the passing of a wave through an object

Explanation:

5 0
2 years ago
A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.
love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0
3 years ago
A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric
Liula [17]

The electric field produced by a single-point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.


1) The first charge is q=-3.00 nC=-3.00 \cdot 10^{-9} C, and it is located at x=0, so its distance from the point x=0.200 m is

r=0.200 m-0=0.2 m

Therefore, the electric field is

E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.


2) The second charge is q=-5.50 nC=-5.5 \cdot 10^{-9}C and it is located at x=0.800 m, so its distance from the point is

r=0.800 m-0.200 m=0.6 m

Therefore, the electric field is

E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.


3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C

and the sign tells us that the field is directed toward negative x-direction.

7 0
3 years ago
The instrument used to measure the diameter of a thin wire is​
Pie

micrometer is used to measure the diameter of a thin wire

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3 years ago
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