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Vanyuwa [196]
2 years ago
10

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

Physics
1 answer:
ikadub [295]2 years ago
8 0

Answer:

b) No acceleration in the vertical

c) 35N

d) 35N

e) 8.75\ m/sec^2

Explanation:

a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book

b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero

c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

Fh_{net}=F-F_r=50 N - 15N=35N

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:

F_n=35N in the x-axis

e) Following Newton's second law, the acceleration is calculated as

a=\frac{F_t}{m}=\frac{35N}{4Kg}=8.75m/sec^2

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Diano4ka-milaya [45]

Answer:

f = 3.1 kHz

Explanation:

given,

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speed of sound = 343 m/s

fundamental frequency  = ?

The fundamental frequency of a tube with one open end and one closed end is,

f = \dfrac{v}{4L}

f = \dfrac{343}{4\times 0.028}

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hence, the fundamental frequency is equal to f = 3.1 kHz

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3 years ago
What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s
Ksivusya [100]

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Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

F=ma

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a is the acceleration

In this problem, we have

m = 90.0 kg

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Substituting, we find the net force on the object:

F=(90.0)(3.0)=270 N

Learn more about Newton's second law:

brainly.com/question/3820012

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Hope this helps! :)

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JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back
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(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second 

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