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Vanyuwa [196]
3 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]3 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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Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

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These three things are:

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Answer:

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Answer:

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Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

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