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ratelena [41]
3 years ago
11

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

Physics
1 answer:
joja [24]3 years ago
6 0

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

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Say you dropped a cannonball from the 17.0–meter mast of a ship sailing at 2.0 meters/second. How far from the base of the mast
Mice21 [21]

The correct choice is A. 0 meters.

If you simply dropped the cannonball and didn't throw it horizontally,

then it'll fall straight down the mast and land on the deck right next to

the mast.


While you were up there holding it, before you dropped it, the cannonball

was moving horizontally, at 2.0 meters/second, along with the rest of the

ship and everything else aboard. It continued doing that after it dropped,

and from the point of view (in the reference frame) of the mast and everyone

on the ship, it fell straight down, parallel to the mast.


Now that we have that question answered, we can proceed to the more-

important ones. I answered the easy one, but YOU'll have to answer these:


==> WHY did you climb the mast carrying a cannonball ?

Have you been drinking sea water or bad rum ?


==> WHY did you drop it, and never even yell "LOOK OUT BELOW !" ?


==> How many formerly-able-bodied souls were injured by the

plummeting cannonball ?


==> What did everybody ELSE yell after the impact ?


==> What did they do to you after they brought you down ?

3 0
3 years ago
Read 2 more answers
What quantities belong in cells X and Y?
satela [25.4K]

Answer:

X: period

Y: tangential speed

Explanation:

100% on quiz your welcome(:

6 0
2 years ago
A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci
Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = \frac{C_{3}C_{4}  }{C_{3} + C_{4}  }

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0
3 years ago
Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different
Sedbober [7]
Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.
5 0
3 years ago
A flat horizontal surface with an area of 518 cm^2 is inside a uniform electric field of 1.33 x 10^4 N/C. The angle between the
GarryVolchara [31]

Answer: 576.48 N*m^2/C

Explanation: In order to calculate the electric flux through the any surface we have to take into account the scalar product between the electric field vector and the normal vector to the surface.

So we have:

ФE= E*A= 1.33 * 10^4*0.0518* cos (33.2°)= 576.48 N*m^2/C

3 0
3 years ago
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