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il63 [147K]
2 years ago
11

Jenna lifts a 5 kg mass through a displacement of 2 m in 3 S. How much time does Kendra need to lift a 10 kg mass through a dist

ance of 1M in order to have the same power output.
Physics
1 answer:
tiny-mole [99]2 years ago
3 0

Power = (work done) / (time to do the work)

Work done = (force) / (distance moved in the direction of the force)

The work for the first mass is (5 kg) x (g) x (2 m) = 10g Joules.

The work for the 2nd mass is (10 kg) x (g) x (1 m) = 10g Joules.

as long as Jenna lifts both masses on the same planet, the amount of work is going to be the same in both cases.  So in order to have the same power output, she would have to do both jobs in the same amount of time.  If she did the first one in 3 seconds, then the second one also requires 3 seconds.

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Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete
disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) F_{net} \text {on wire }3=0

\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm

position of wire = 50 - 1.2

= 48.8cm

b)  F_{net} \text {on wire }1=0

\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A

Direction ⇒ downward

5 0
3 years ago
When an object falls, it trades gravitational potential energy for kinetic energy, accelerating toward the ground. calculate the
liq [111]
The speed of the energy decreases
5 0
3 years ago
How to reduce the energy loss from the study lamp?​
AnnZ [28]

Stop using it all the time for some useless things.

7 0
3 years ago
What is the relationship between the acceleration of an object change in relation to its mass
yarga [219]

The relationship between mass and acceleration is an inverse proportionality

Explanation:

The relationship between the acceleration of an object and its mass is given by Newton's second law, which states that:

F=ma

where

F is the net force on the object

m is its mass

a is its acceleration

From the equation, we notice that if the force on the object is kept constant, then the mass and the acceleration are inversely proportional to each other. This means that:

  • If the mass of the object is increased, its acceleration will decrease
  • If the mass of the object is decreased, its acceleration will increase

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

3 0
3 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

8 0
2 years ago
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