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Nikitich [7]
3 years ago
9

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction? A0.01 B0.20 C0.38 D0.48
Physics
2 answers:
k0ka [10]3 years ago
5 0
The formula is Force = coefficient of friction times the weight or normal force (in newtons). Plug the numbers in the formula. Do the simple division math. The coefficient of friction should be any number between 0 and 1. If your answer is smaller than 0 or greater than 1, go back and check your work. The answer for this question is C. 0.38   
baherus [9]3 years ago
3 0

Answer:

\mu_k = 0.38

Explanation:

As we know that the normal force on the book is given as

F_n = 51.7 N

also the force required to push the book on the table is given as

F_f = 19.5 N

now by the formula of friction force we know that

F_f = \mu_k F_n

now plug in all values in it

19.5 = \mu_k (51.7)

here we will have

\mu_k = \frac{19.5}{51.7}

\mu_k = 0.38

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sdas [7]

In spring mass system we know that angular frequency is given as

\omega = 2\pi f

f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

now we know that speed of SHM at its extreme position is given by

v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

7 0
3 years ago
a lawn mower is pushed with a force of 50n. if the angle between the handle of the mower and the ground is 30°. why doesn't the
Mademuasel [1]

Answer:

IT doesnt push down because the lawnmower has wheels and the ground is hard

Explanation:

Branliest

5 0
3 years ago
If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula
snow_lady [41]

Here we know that

\omega_i = - 8.4 rad/s

\alpha = - 2.8 rad/s^2

t = 1.5 s

now from kinematics we have

\omega_f = \omega_i + \alpha t

now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

\omega_f = -8.4 + (-4.2)

\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

6 0
3 years ago
An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17
Softa [21]

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

3 0
3 years ago
A 1200 kg sports car accelerates from 0 m/s to 30 m/s in 10 s. What is the average power of the engine?
DIA [1.3K]

Answer:

3600N

Explanation:

Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s

ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

5 0
2 years ago
Read 2 more answers
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