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pashok25 [27]
1 year ago
12

A pulley system used to lift car tires has a mechanical advantage of 11.2. If you pull on the pulley with a force of 150 N, how

much force does the pulley exert on the tire?
Physics
2 answers:
Novay_Z [31]1 year ago
7 0
1680 is the answer i got because i did this problem in apex
IgorC [24]1 year ago
5 0

Answer:

Load = 1680 N

Explanation:

Mechanical advantage of the pulley system, MA = 11.2

If a person pull on a pulley with a force of 150 N we have to find the amount of force the pulley exert on the tire. Mechanical advantage of a pulley system is given by the ratio of load to the effort force i.e.

IMA=\dfrac{load}{effort}

11.2=\dfrac{load}{150\ N}

Load = 1680 N

Hence, the force the pulley exert on the tire is 1680 Newtons.

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Shareen performs a skit to model a method of charging. In the skit, a painter shakes her hand and gets paint on her.
Veseljchak [2.6K]

Answer:

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

Explanation:

Conduction phenomenon of charging is the process of charging in which two bodies are made in contact with each other so that charges are transferred due to potential difference of two bodies.

here we know that when hands are shake then it will have paint on it. so here due to hand shake the hands are in contact with charge particles and due to contact the electrons are transferred to the hand.

Now here we need to assume that charge of paint must be opposite that of the charge on the hand because only due to opposite charge attraction the paint must be transferred to the hand

SO here correct answer will be

It models conduction because the painter represents a charged object and the paint represents electrons that are transferred through contact.

5 0
1 year ago
Read 2 more answers
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
Let’s now multiply two numbers in scientific notation using Google.
olga_2 [115]
We are going to rewrite each number:
 (4.48E-8) = 0.0000000448
 (5.2E-4) = 0.00052
 We observe that when multiplying, the exponent will be on the order of 10 ^ -11
 Doing the multiplication we have:
 (4.48E-8) * (5.2E-4) = 2.3296E-11
 Rewriting:
 (4.48E-8) * (5.2E-4) = 2.33E-11
 Answer:
 
2.33E-11
5 0
2 years ago
Read 2 more answers
Assume that a cloud consists of tiny water droplets suspended (uniformly distributed,
aev [14]
9.8 ms^-2 is acceleration
4 0
2 years ago
A tube with a cap on one end, but open at the other end, produces a standing wave whose fundamental frequency is 130.8 Hz. The s
Angelina_Jolie [31]

Answer:

A. 261.6 hz.

B. 0.656 m.

Explanation:

A.

When yhe tube is open at one end and closed at the other,

F1 = V/4*L

Where,

F1 = fundamental frequency

V = velocity

L = length of the tube

When the tube is open at both ends,

F'1 = V/2*L

Where

F'1 = the new fundamental frequency

Therefore,

V/2*L x V/4*L

F'1 = 2 * F1

= 2 * 130.8

= 261.6 hz.

B.

F1 = V/4*L

Or

F'1 = V/2*L

Given:

V = 343 m/s

F1 = 130.8

L = 343/(4 * 130.8)

= 0.656 m.

8 0
1 year ago
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