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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.

You might be interested in
A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68
liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0
3 years ago
Read 2 more answers
In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir
Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be S_{1}

and the distance covered by Sir Alfred be S_{2}

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

S = ut + \frac{1}{2}at^{2}

Where S is the distance traveled

u is the initial velocity

a is the acceleration

and t is the time

For Sir George,

S = S_{1}

u = 0 m/s (Since they start from rest)

a =0.21 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{1}  = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1}  = 0.105 t^{2}\\

t^{2} = \frac{S_{1}}{0.105}

Now, for Sir Alfred

S = S_{2}

u = 0 m/s (Since they start from rest)

a =0.26 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{2}  = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2}  = 0.13 t^{2}\\

t^{2} = \frac{S_{2}}{0.13}

Since, they traveled for the same time, t just before collision, we can write

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

S_{1} + S_{2} = 86 m

Then, S_{2} = 86 - S_{1}

Then,

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13} becomes

\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}

0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}

S_{1} = 38.43 m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0
3 years ago
There are several ways to model a compound one type of model is shown ?what is the chemical formula for the molecule modeled?
s2008m [1.1K]

The is organic compound with the correct chemical formula C4H9O2.

<h3>What is a model?</h3>

A model is a representation of reality. A model serves the purpose of prediction as well as explanation.

Looking at the model of the molecule we can see that it is the organic compound with the correct chemical formula C4H9O2. The molecule is shown in the image attached to this answer.

Missing parts:

There are several ways to model a compound. One type of model is shown.

What is the chemical formula for the molecule represented by the model?

CHO

C4H9O2

C4H8O

C3H8O2

Learn more about molecular models:brainly.com/question/156574?

#SPJ1

7 0
2 years ago
An airplane flies 1350km in 2 1/4 hours. What is its average speed in kilometers per hour?
Brut [27]
600km since it would be 1350 divided by 2.25
8 0
3 years ago
for a body moving in a straight line / its distance and displacement can be same. justify with an example?
Tema [17]
The distance of an object is the total distance it travels, while the displacement of the object is how far away the object is from the starting point.

Because the body is moving in a STRAIGHT line, that means it does not change directions, therefore when the body gets to the destination, the total distance will be the same as the displacement. If the body were to change directions, then the magnitude of the vectors will need to be added up and calculated.

For example, let's say you are walking to your friends house directly across the street from your house. All you need to do is walk in a strsight line from the front of your house and you will get to your friends house. The distance and displacement will be the same.

Now if your friend lived right across the street, but 5 houses down, and you cross the street directly from your house then turn in another direction and walk straight, then the distance in this case will quite likely be greater than the displacement because the displacement is the distance from your house to your friends house when measured diagonally.
6 0
2 years ago
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