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egoroff_w [7]
3 years ago
15

the other end of the pipe. With the aide of the pipe, does the applied force produce a smaller torque, a greater torque, or the

same torque on the nut?
Physics
1 answer:
pav-90 [236]3 years ago
3 0

Answer:

With the addition of the pipe we have a greater torque.

Explanation:

We need to complete the description of the problem, searchin in internet we have:

"Sometimes, even with a wrench, one cannot loosen a nut that is frozen tightly to a bolt. It is often possible to loosen the nut by slipping one end of a long pipe over the wrench handle and pushing at the other end of the pipe. With the aid of the pipe, does the applied force produce a smaller torque, a greater torque, or the same torque on the nut?"

With the addition of the pipe we have a greater torque, as it increases the distance or radius of torque.

We know that torque is defined, as the product of force by distance, in this way we have:

T = F * d

where:

T = torque [N*m]

F = force [N]

d = distance [m]

We can see in the above equation, that increasing the distance increases torque proportionally.

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Pressure that increases with depth in a swimming pool is called ______________ pressure.
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4 0
4 years ago
What is the car’s average velocity (in m/s) in interval between t=1.0s to t=1.5s?
Lemur [1.5K]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula for velocity is;

Velocity (in m/s) = distance/time

The distance the car covered in the completed question is divided by the difference in the time interval

The difference in the time interval will be = 1.5s - 1.0s = 0.5s

NOTE: the distance must be in meters or be converted to meters

7 0
3 years ago
The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
jek_recluse [69]

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

4 0
4 years ago
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