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Svet_ta [14]
3 years ago
15

A 5.00-kg box slides 6.50 m across the floor before coming to rest. what is the coefficient of kinetic friction between the floo

r and the box if the box had an initial speed of 3.00 m/s?
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0
Work=Fd and is equal to the change in energy, friction does work to stop the box. The box's kinetic energy is initially (K=.5m(v^2)) 22.5J and ends up at 0J. Friction does 22.5J of work over 6.5m. 22.5J=(F)(6.5m), F=3.46N. The force of friction is equal to the coefficient of friction times the normal force of an object, F=μFⁿ. We can determine the normal force by multiplying mass and gravity, Fⁿ=50N. Bringing this back to the force applied by friction, 3.46N/50N=μ=0.0692.
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Are you a negro can i call you one
Roman55 [17]

Answer:

no it is mean call people their names

Explanation:

because if my name was Shar i would want people to call me Shar not negro

3 0
2 years ago
A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor
Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is  located on the x-axis=\lambda_1

Linear charge density of another long line which is  located on the y-axis=\lambda_2

We have to find the direction of electric field at z=a on the positive z-axis if \lambda_1 and \lambda_2 are positive.

The direction of electric field  at z=a on the positive z-axis  is positive z-direction .

Because \lambda_1 and \lambda_2 are positive and the electric field is  applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

6 0
3 years ago
A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o
aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

E_r = \frac{1}{2} mv^2

Substitute,

14kg for m

9m/s for v

E_r = \frac{1}{2} (14) (9)^2\\= 567J

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

E_R = \frac{1}{2} Iw^2

The expression for the moment of inertia of the cylinder is,

I = \frac{1}{2} mr^2

The expression for angular velocity is,

w = \frac{v}{r}

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2

= \frac{mv^2}{4}

E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

E = E_r + E_R\\\\

substitute 567J for E(r) and 283.5J for E(R)

E = 567J + 283.5\\= 850.5J

The total energy of the cylinder is 850.5J

6 0
3 years ago
Read 2 more answers
Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,
andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

<u>D = - 0.5i - 0.25j</u>

Therefore,

<u>Dx = -0.5</u>

<u>Dy = -0.25</u>

8 0
3 years ago
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masya89 [10]
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3 0
3 years ago
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