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evablogger [386]
3 years ago
13

An object's mass refers to _____ and an object's weight refers to _____. Fill in each blank.

Physics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:

c

Explanation:

because the more mass it has the more dense it is

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I need help this is due today GIVING BRAINLIEST IM DESPERATE!
Nitella [24]

Answer:

1: is A

2: is C

Explanation:

1: because all other option would imply that the truck was not moved from from its previous position due to the lack of work / force exerted by the car

2: this is because the less dense ball will have less opposing force action on it causing it to move / accelerate / go at a faster speed

3 0
3 years ago
How does a dynamo Work?
marin [14]
The generator/dynamo<span> is made up of stationary magnets (stator) which create a powerful magnetic field, and a rotating magnet (rotor) which distorts and cuts through the magnetic lines of flux of the stator. When the rotor cuts through lines of magnetic flux it makes electricity.</span>
8 0
3 years ago
Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged
MArishka [77]

Answer:

A)   E = 0N/C

B)   0i + 0^^j

C)   F = 0N

D)   0^i  + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

E=k\frac{rQ}{(r+R^2)^{3/2}}               (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

F=qE=q(0N/C)=0N

D) The direction of the force is 0^i + 0^j

6 0
3 years ago
The index of refraction for red light in water is 1.331 and that for blue light is 1.340. If a ray of white light enters the wat
kobusy [5.1K]

Answer:

Angle of refraction for red light is 43.01^{\circ}

Angle of refraction for blue light is  42.68^{\circ}    

Explanation:

It is given refractive index for red light is \mu _{red}=1.331

Refractive index of blue light \mu _{blue}=1.340

Angle of incidence i=65.30^{\circ}

According to law of refraction \mu =\frac{sini}{sinr}

For red light 1.331 =\frac{sin65.30^{\circ}}{sinr}

1.331 =\frac{0.908}{sinr}

sinr=0.682

r=43.01^{\circ}

Therefore angle of refraction for red light is 43.01^{\circ}

Similarly for blue light 1.340 =\frac{sin65.30^{\circ}}{sinr}

1.340 =\frac{0.908}{sinr}

sinr=0.677

r = 42.68^{\circ}

Therefore angle of refraction for blue light is  42.68^{\circ}

6 0
3 years ago
Guys help me please gr5 science pls help
Stolb23 [73]

Answer:

just google it.

Explanation:

Because its easier

7 0
4 years ago
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