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ss7ja [257]
2 years ago
8

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

at is the friction force you must overcome to push the cart up the ramp
Physics
1 answer:
pentagon [3]2 years ago
4 0

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

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Short-range forecasts tends to ________ longer-range forecasts.
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What body in the solar system do you think is one focus of the moons orbit
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2 years ago
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In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and
Savatey [412]

<u>Answer:</u>

<em>Newtons II law: </em>

<em>     </em>It  is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

<em>For the hanging weight/mass</em>

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

           Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

        Now, Fnet = m₁ .g - T

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So From Newtons II law<em> F =  m.a</em>

3 0
2 years ago
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A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.
Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

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b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0
2 years ago
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