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Dmitrij [34]
3 years ago
9

PLEASE HELPPPPP I WILL GIVE POINTS AND BRAINIEST

Physics
2 answers:
Elza [17]3 years ago
5 0
I agree^ fjidnsnwjskocucuhrebnaosicuej
UNO [17]3 years ago
3 0

Answer:

Graph A

Explanation:

I hope this is right

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Which factor is most important to keep the same when comparing the<br> hardness of solids?
svet-max [94.6K]

Answer:

i think B

Explanation:

just a guess

8 0
4 years ago
How many ohms of resistance are in a 120–volt hair dryer that draws 7.6 amps of current?
miv72 [106K]

From Ohm's law . . . Resistance = (voltage) / (current)

Resistance = (120 volts) / (7.6 Amperes)

<em>Resistance = 15.8 Ω</em>

7 0
3 years ago
Days when the sun’s light hits both hemispheres equally are called __________.
anyanavicka [17]
It is called an equinox. This days happen because the plane of the Earth's equator passes through the center of the Sun, in result, the light coming from it hits both hemispheres equally. Hopefully my answer has come to your help.
8 0
3 years ago
A 2.0 kg block is given an initial speed of 8.0 m/s up an incline plane of angle 30° where the coefficient of friction is 0.35.
Korvikt [17]

Answer:

4.1 m

Explanation:

Given :

Mass of the block = m = 2 kg.

Initial velocity = v_{i} = 8 m/s

Angle of the incline = α = 30°

Coefficient of friction = μ = 0.35

Distance moved up the incline is calculated using the work energy theorem.

Work done by the net force =  change in kinetic energy of the object.

Net work = work done by friction + work done by the gravity component.

(- mg sin 30 - μ mg cos 30 ) d = 1/2m v_{f}^{2} - 1/2 m v_{i}^{2}

m cancels out when divided on both sides with m.

- [(9.8 sin 30 - ( 0.35  × 9.8 × cos 30) ] d = 1/2 ( 0² - 8² )

⇒ -7.87 d = -32

⇒ Distance traveled up the incline = d = 4.0658 m = 4.1 m

8 0
4 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
3 years ago
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