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Thepotemich [5.8K]
3 years ago
12

Lesson Check

Physics
1 answer:
disa [49]3 years ago
4 0
The answer should be B
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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
3 years ago
Read 2 more answers
Which of these are part of our solar system? select all that apply
Len [333]
A)planets
b)the sun
c)moons
e)comets
f)asteroids
6 0
3 years ago
Read 2 more answers
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
3 years ago
7.1 Project Guidelines 2021
Ede4ka [16]
I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and
3 0
3 years ago
a car initially at 65.00 m/s accelerates at 22.39 m/s^2. How far has the car traveled before it comes to a stop
Gennadij [26K]

Answer:

94.35 meter

Step by step explanation

7 0
2 years ago
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