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Phantasy [73]
2 years ago
7

A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N, directed 30 degrees below the

horizontal.
a.) Calculate the force of friction acting on the box.
b.) Calculate the normal force acting on the box.
Physics
1 answer:
alina1380 [7]2 years ago
3 0

Answer:

Explanation:

Assuming the table is horizontal

As speed is constant, friction must exactly balance the horizontal component of the applied force

Ff = 50cos30 = 43 N

The  normal force must oppose both the weight of the box and the vertical component of the applied force.

N = 10(9.8) + 50sin30 = 123 N

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Consider the three displacement vectors
matrenka [14]

Answer:

Explanation:

\overrightarrow{A} = 3\widehat{i}+3\widehat{j}

\overrightarrow{B} = \widehat{i}-4\widehat{j}

\overrightarrow{C} = -2\widehat{i}+5\widehat{j}

(a)

\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}

\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}

Magnitude of \overrightarrow{D} = \sqrt{2^{2}+4^{2}}

                                                                     = 4.47 m

Let θ be the direction of vector D

tan\theta =\frac{4}{2}

θ = 63.44°

(b)

\overrightarrow{E} = - \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}

\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}

Magnitude of \overrightarrow{E} = \sqrt{6^{2}+6^{2}}

                                                                     = 8.485 m

Let θ be the direction of vector D

tan\theta =\frac{6}{-6}

θ = 135°

4 0
4 years ago
How does the input distance of a single fixed pulley compare to the out- put distance?
ololo11 [35]

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

<u>Explanation:</u>

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

4 0
3 years ago
a motorcar is moving with a velocity of 108 km / h and it takes 4s to stop after the brakes are applied calculate the force exer
k0ka [10]

Answer: - 7500N

Explanation:

Given the following :

Initial Velocity of car = 108km/hr

Time taken to stop after applying brakes = 4s

Mass of passengers in car = 1000kg

Force exerted by the brakes on the car =?

After 4s, then final Velocity (V) = 0

Initial Velocity (u) of the car = 108km/hr

108km/hr = (108 × 1000)m ÷ (3600)s = 30m/s

Force exerted = mass(m) × acceleration(a)

Acceleration of car = Change in Velocity with time

a = (v - u) / t

a = (0 - 30) / 4

a = - 30/ 4

a = - 7.5m/s^2

Therefore,

Force exerted = mass(m) × acceleration(a)

Force exerted = 1000kg × (-7.5)m/s^2

Force exerted = - 7500N

5 0
3 years ago
Read 2 more answers
with what speed does a freely falling object dropped from a height of 88.2m hit the ground? how long does it take for it hit the
Kryger [21]

Answer:

Vf = 41.6 [m/s].

Explanation:

To solve this problem we must use the equations of kinematics.

Vf² = Vo² + (2*g*y)

where:

Vf =  final velocity [m/s]

Vo = initial velocity = 0

g = gravity acceleration = 9.81 [m/s²]

y = height = 88.2 [m]

Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.

Vf² = Vo² + (2*g*y)

Vf² = 0 + (2*9.81*88.2)

Vf = (1730.48)^0.5

Vf = 41.6 [m/s]

6 0
3 years ago
Beatrice and the Elevator Beatrice, a middle school student, is visiting a very tall office building and notices that she feels
ahrayia [7]

Explanation:

The question is not complete, here is the complete question

<em>"Beatrice and the Elevator Beatrice, a middle school student, is visiting a very tall office building and notices that she feels heavier when the elevator car is traveling up and lighter when the elevator car is traveling down. After making these observations, Beatrice comes back to the building and stands on a bathroom scale that measures her weight as she travels up and down in the elevator.</em>

<em>1. . What question is Beatrice trying to answer?</em>

<em>2. What is one variable Beatrice could change in her investigation? What might she figure out if this </em>

<em> variable was changed"</em>

1. Beatrice is trying to observe the influence of the elevator movement on her weight, Hence the question is<em> "will the elevator  movement cause her weight to change"</em>

therefore moving upward the reading on the scale will increase

Reading=mg+ma

downward

Reading will reduce

Reading=mg-ma

2. The independent variable is the acceleration due to gravity g=9.81m/s^2

while the dependent variables are

    i. The elevators acceleration

    ii. Beatrice's mass

6 0
3 years ago
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