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Mariana [72]
3 years ago
12

A sled is pulled with a horizontal force of 18 n along a level trail, and the acceleration is found to be 0.39 m/s2. an extra ma

ss m = 4.5 kg is placed on the sled. if the same force is just barely able to keep the sled moving, what is the coefficient of kinetic friction between the sled and the trail?
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0
From Newton's second law of motion, it is identified that the net force applied to the object with mass m, will make it move with an acceleration of a. This can be mathematically translated as,
                        F = m x a
To solve for the mass of the sled, we derive the equation above such that,
                        m = F / a
Substituting,
                       m = (18 N) / (0.39 m/s²)
                          m = 46.15 kg

Then, we add to the calculated mass the mass of the extra material.
                      total mass = 46.15kg + 4.5 kg
                       total mass = 50.65 kg
We solve for the normal force of the surface to the object by calculating its weight.
                     F₂ = (50.65 kg)(9.8 m/s²)
                     F₂ = 496.41 N

The force that would allow barely a movement for the object is equal to the product of the normal force and the coefficient of kinetic friction.
                     F = (F₂)(c)
                      c = F/F₂

Substituting,
                      c = 18 N/496.41 N
                       c = 0.0362

<em>ANSWER: c = 0.0362</em>

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Explanation:

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4 0
2 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 8.0 s, it rotates 35 rad.
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Answer:

(a) 1.093 rad/s^2

(b) 4.375 rad/s

(c) 8.744 rad/s

(d)  67.845 rad

Explanation:

initial angular velocity, ωo = 0

time, t = 8s

angular displacement, θ = 35 rad

(a) Let α be the angular acceleration.

Use second equation of motion for rotational motion

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

35 = 0 + 0.5 x α x 8 x 8

α = 1.093 rad/s^2

(b)  The average angular velocity is defined as the ratio of total angular displacement to the total time taken .

Average angular velocity = 35 / 8 = 4.375 rad/s

(c) Let ω be the instantaneous angular velocity at t = 8 s

Use first equation of motion for rotational motion

ω = ωo + αt

ω = 0 + 1.093 x 8 = 8.744 rad/s

(d) Let in next 5 seconds the angular displacement is θ.

\theta =\omega _{0}t+\frac{1}{2}\alpha t^{2}

By substituting the values

θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5

θ = 67.845 rad

8 0
3 years ago
100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.
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V=13014.1 V
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P=VI
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I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A

b) The voltage change along each line can be found by using Ohm's law:
\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V

c) The power wasted as heat along each line is given by:
P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W
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6 0
3 years ago
A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red
Colt1911 [192]

Answer:

6.59\times 10^7m/s

Explanation:

We are given that

Wavelength of red light=\lambda=635nm

Wavelength of  green light=\lambda'=510 nm

We have to find the speed of physicist traveling according to his own testimony.

We know that

\frac{\lambda}{\lambda'}=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}

Where c=3\times 10^8m/s

Substitute the values

\frac{635}{510}=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}

1.25=\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}

Squaring on both sides

1.5625=\frac{1+\frac{v}{c}}{1-\frac{v}{c}}

1.5625-1.5625\frac{v}{c}=1+\frac{v}{c}

1.5625-1=\frac{v}{c}+1.5625\frac{v}{c}=\frac{v}{c}(1+1.5625)

0.5625=2.5625\frac{v}{c}

v=\frac{0.5625c}{2.5625}=\frac{0.5625\times 3\times 10^8}{2.5625}

v=6.59\times 10^7m/s

8 0
3 years ago
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