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Xelga [282]
3 years ago
10

Help me with this plsss

Physics
1 answer:
Artist 52 [7]3 years ago
8 0

Answer:

that is going at a constant rate

Explanation:

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I am no use to you i forgot this concept
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When describing how power and work are similar you would want to mention that a you must know time to determine both work and po
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The answer that would best complete the given statement above would be option B. When describing how power and work are similar you would want to mention that <span>you must know force in order determine both work and power. Hope this answers your question. Have a great day!</span>
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4 years ago
Read 2 more answers
1. (6x + 8)(5x - 8)
olga55 [171]

Answer:

form 1 question??????????

7 0
3 years ago
if the current through a resistor is increased by a factor of 4, how does this affect the power dissipated?
lbvjy [14]

The new current will be 4I. The power dissipated by the resistor will increase by a factor of 16.

<h3>What is a resistor?</h3>
  • Using electrical resistance as a circuit element, a resistor is a passive electrical component with two terminals. In electrical circuits, resistors are used, among other things, to limit current flow, modify signal levels, divide voltages, bias active devices, and terminate transmission lines.
  • As test loads for generators, power distribution systems, and motor controls, high-power resistors that can create many watts of heat rather than just electrical energy can be used.
  • Variable resistors can be used as sensors for force, heat, light, humidity, humidity, and chemical activity or for adjusting circuit components.
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brainly.com/question/24858512

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4 0
1 year ago
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i
bija089 [108]

Answer

Mass m = 78 kg

Vertical height in each stage h = 11 m

(a).

Initial speed u = 0

Final speed v = 1.1 m / s

v^2=u^2 + 2 as

1.1^2 = 2 a \times 11

a = 0.055 m/s²

Work done

W_a= m g h + \dfrac{1}{2}mv^2

W_a= 78\times 9.8 \times 11 + \dfrac{1}{2} 78 \times 1.1^2

W_a = 8408.4 + 47.19

W_a = 8455.59 J

(b).Work done

W_b= mgh

W_b = 78× 9.8× 11

W_b= 8408.4 J

c)

Work done

W= m g h + \dfrac{1}{2}m(v_f-v_i)^2

Where V = final speed

               = 0

            v = 1.1 m / s

for deceleration a = -0.055 m/s²

now,

F_L = 56 (-0.055+9.8) = 545.72 N

W_c = 545.75 × 11

W_c = 6003.25 J

   

7 0
3 years ago
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