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Paha777 [63]
3 years ago
8

A system of pulleys is used to raise a load of bricks that weighs 1,700 newtons. The force applied to the pulley is 340 newtons.

What’s the mechanical advantage of this machine? 1 2 5 10
Physics
2 answers:
Alchen [17]3 years ago
6 0

Answer:

The answer is 5

Explanation:

I got this question on a test and this was the right answer

Please mark as brainliest :)

Law Incorporation [45]3 years ago
4 0
By definition, we have that the mechanical advantage is given by the following equation:
 MA =  \frac{W}{T} 

 Where,
 W: is the load
 T: is the tension
 Substituting the values in the given equation we have:
 MA = \frac{1700}{340}
 MA = 5

 Therefore, the mechanical advantage is equal to 5.
 Answer:
 The mechanical advantage of this machine is:
 MA = 5
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8 0
3 years ago
Which of the following should a warm up NOT include?
Arlecino [84]
The answer to this is ax
7 0
3 years ago
Tory has a mass of 40kg. She sleds down a hill that has a slope of 25 degrees. what is the component of her weight that is along
Fudgin [204]

1.7 x 10^2 N

or 166 N

First you find the vertical component of the weight, which is 9.8*40, (g*m), which is 392 N. You then find the angle between that and the slope, which is 90-25, which is 65. You then multiply the vertical weight by cos(65), to find the component of that that is parallel to the slope. You get 165.666 N

3 0
3 years ago
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?
Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

  • the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

        Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

  • the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

         Q = 0.456 x 33.5 x 10^(4) = 152760 J

  • the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

        Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0
3 years ago
A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-1
VARVARA [1.3K]

Answer:

\theta=40^0

Explanation:

The magnitude of the magnetic force is

F_m=evB\sin\theta

To find the angle, we make \sin\theta subject of the formula

\implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}

\implies \sin\theta=0.641025641

\therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0

8 0
3 years ago
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