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torisob [31]
3 years ago
5

3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration

(in m/s2) down the ramp? (Enter the magnitude.) m/s2 What is the force (in N) of the ramp on the block? (Enter the magnitude.) N (b) What force (in N) applied upward along and parallel to the ramp would allow the block to move with constant velocity?
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

a)  a = 7.72 m / s²,  N = 19.9 N  and b)   F = 25.5 N

Explanation:

To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

    sin52 = Wx / W

    cos52 = Wy / W

    Wx = W sin52

    Wy = w cos 52

Let's write them equations

X axis

    Wx = ma

Y Axis

    N-Wy = 0

    N = Wy

a) Let's calculate the acceleration

    a = W sin52 / m = mg sin 52 / m

    a = g sin 52

    a = 9.8 sin52

    a = 7.72 m / s²

The force of the ramp is normal

    N = Wy = mg cos 52

    N = 3.3 9.8 cos 52

    N = 19.9 N

b) For the block to move at constant speed the sum of force on the axis must be zero,

    F - Wx = 0

    F = Wx

    F = mg sin52

   F = 3.3 9.8 sin 52

   F = 25.5 N

Parallel to the plane and going up

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A 12 kg block is accelerating at the rate of 9.0 m/s? while being acted on by two forces.
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PLEASE ANSWER, I NEED HELP
Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is 1.56\cdot 10^{-3} N

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

1)

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 47 kg is the mass of Ellen

m_2 = 7.35\cdot 10^{22} kg is the mass of the moon

r=3.84\cdot 10^8 m is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N

2)

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

a=\frac{F}{m}

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0
3 years ago
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