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joja [24]
4 years ago
5

Please help. I got behind and I don't know this.

Physics
2 answers:
enyata [817]4 years ago
5 0
I can help you, just me the question :)
Luden [163]4 years ago
4 0
Dynamic
Contact
Impact
Regional
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Water vapor in earth's atmosphere primarily absorbs which type of photon
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What does the doppler effect indicate about the motion of galaxies?
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4 years ago
A 1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in
Talja [164]

Answer:

132 N

Explanation:

Given that a  1.1 kg hammer strikes a nail. Before the impact, the hammer is moving at 4.5 m/s; after the impact it is moving at 1.5 m/s in the opposite direction. If the hammer is in contact with the nail for 0.025 s, what is the magnitude of the average force exerted by the hammer on the nail

From Newton 2nd law of motion,

Change in momentum = impulse.

Change in momentum = m( V - U )

Substitute all the parameters into the formula

Change in momentum = 1.1 ( 4.5 - 1.5 )

Change in momentum = 1.1 × 3

Change in momentum = 3.3 kgm/s

Impulse = Ft

That is,

Ft = 3.3

Substitute time t into the formula above

F × 0.025 = 3.3

F = 3.3 / 0.025

F = 132 N

Therefore, the magnitude of the average force exerted by the hammer on the nail is 132 N.

3 0
4 years ago
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a po
Archy [21]

Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

(c) ΔU = 5.204nJ

Explanation:

a)

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

6 0
3 years ago
Can someone help me with this please...It's Earth Space Science
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Your answer is correct: Resident 4 would provide the scientist with the most valid results.
3 0
3 years ago
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