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Black_prince [1.1K]
3 years ago
11

A type of wave that carries energy from one place to another, even through

Physics
1 answer:
Igoryamba3 years ago
8 0

Answer:

light waves :)

Explanation:

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A proton is released in a uniform electric field, and it experiences an electric force of 2.04×10−14 N toward the south.
Lemur [1.5K]

Answer:

Part A:

E=127500N/C\\E=1.275*10^{5} N/C

Part B:

Option B (Towards the South)

Explanation:

Part A:

Magnitude if electric field E:

E=Force/charge

Force=2.04×10−14 N

Charge=1.6×10−19 C

E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C

Part B:

Option B (Towards the South)

As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.

7 0
3 years ago
Which of the following graphs represents constant positive acceleration
nika2105 [10]
W, because as time is moving up at a consistent rate the speed is as well, creating the straight line.
5 0
3 years ago
Which statement about oceans is incorrect?
Hitman42 [59]
It is C because less than one percent of water is fresh water
7 0
3 years ago
A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of
Citrus2011 [14]

Answer:

a) (Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\

The buoyancy force acting on the balloon is:

Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0
3 years ago
You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what
Tatiana [17]
27.9 idkkkk look it up on photomath
8 0
3 years ago
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