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wolverine [178]
3 years ago
6

What is the voltage of a computer with 30 Ω of resistance and 15 amps of current?

Physics
1 answer:
Kryger [21]3 years ago
5 0

Answer:

120 V

Explanation:

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When a comet is within the inner solar system, its visible tails point __________.
astraxan [27]

away from the sun

hope it helps...!!!!

6 0
2 years ago
Two stones are dropped from the edge of a 60m cliff , the second stone 1.6secon after the first . How far below the top of the c
tigry1 [53]

Answer:

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let<em> t</em> represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 =  (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t²  = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 =  ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

5 0
3 years ago
Why do astronauts muscles weaken?
Pavlova-9 [17]
They work on a weightless planet, therefore they have very little muscle control since they float in space !!
3 0
4 years ago
A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30
Ivan

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c)  1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed  when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis )  ----- ( 1 )

f ( friction force )=  ( 16400v * 6.25 *10^-2) =  1025 v

F ( force applied ) = 75

W ( weight of  block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = \frac{75- (98.1*sin30^{0}) }{1025} = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = \frac{0 - (98.1*sin30^{0}) }{1025} =  - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = \frac{75+( 98.1*sin30^{0})  }{1025} = 1.21 * 10^-1 m/s

5 0
3 years ago
Number 25 and 26 please!!!
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laborious ... enclosed may help ... cos -angle = cos +plus angle ... sin - anfle = - sine angle

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4 years ago
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