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Yuki888 [10]
3 years ago
12

Three forces are exerted on an object placed on a slope in the figure. The forces are measured in Newton (N).

Physics
2 answers:
kobusy [5.1K]3 years ago
4 0

Answer:

1. Component = F - mgsin\theta - \mu R

2. Normal

3. F_{up} - F_{friction}

4 . upwards

Explanation:

1. The diagram on a slope shows the forces acting on an object. There a three sets of forces that are on the object. One is the normal force, N. The other one is the pulling force, which shall be called .

The last component is the force acting on the object due to friction. this is given as F_{total}  = mgsin\theta

2. The perpendicular force is the normal. It is given as N

3. The magnitude of the Fnet is the difference in all the forces.

4. Overall, the net force will be upwards as the object is being pulled in that direction.

Vika [28.1K]3 years ago
3 0
<span>The component of the net force that is parallel to the slope is the friction force. It is the force due to the friction present between the object and the surface of the slope. The force perpendicular to the slope is called the normal force and it is the contact force exerted by the object to the surface of the slope.</span>
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WARRIOR [948]

Given Information:

Mass of elephant = m = 750 kg

Height = h = 14.3 m  

time = t = 30 seconds

Required Information:

Power needed to lift elephant = P = ?

Answer:

Power needed to lift elephant ≈ 3507 watts

Explanation:

As we know power is given by

P = PE/t

Where PE is the potential energy and t is the time

Potential energy is given by

PE = mgh

Where m is the mass of elephant, g is the gravitational acceleration and h is the height to lift the elephant.

PE = 750*9.81*14.3

PE = 105212.25 Joules

Therefore, the required power to lift the elephant is

P = PE/t

P = 105212.25/30

P ≈ 3507 watts

8 0
4 years ago
Please need help on this
mote1985 [20]

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-- The density of anything never depends on how much of it you have.

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--  Choose ' <em>A </em>' .

3 0
3 years ago
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PLEASE HELP ASAP!!!<br> GIVING BRAINLIEST!!!<br> 40 points!!
Nana76 [90]
W = Fd = 4(2100) = 8400 J

So the answer is A) 8400 J


I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)

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4 0
3 years ago
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A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from
Rzqust [24]

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

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7 0
2 years ago
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A climber pulls down on a rope causing his body to rise up with the rope? Which law of motion is it?
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3 years ago
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