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3 years ago

15

crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50. A 20-N force

is applied to the crate acting to the right. What is the resulting static friction force acting on the crate

1 answer:

mr_godi [17]3 years ago

6 0

Answer:

The resulting static friction force acting on the crate is 10 N.

Explanation:

Given that,

Coefficient of static friction = 0.50

Normal force = 20 N

We need to calculate the resulting static friction force acting on the crate

Using formula of static friction force

$f_{s}=\mu_{s}N$

Where, N = normal force

$\mu_{s}$ = Coefficient of static friction

Put the value into the formula

$f_{s}=0.50\times20$

$f_{s}=10\ N$

Hence, The resulting static friction force acting on the crate is 10 N.

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A 36-N crate is suspended from the left end of a plank. The plank weighs 21 N, but it is not uniform, so its center of gravity d

riadik2000 [5.3K]

Answer:

0.51 m

Explanation:

from the question we are given the following

weight of the crate = 36 N

weight of the plank = 21 N

distance of of the balance from the left end = 0.3 m

from the diagram attached, CG is the center of gravity ( the point near or within a body through which its weight can be assumed to act ), taking the CG as where the weight of the plank acts and it being at a distance L from the support

we can find the distance of the center of gravity just as we would find the moment about the support balance

therefore

36 x 0.3 = 21 x L

10.8 = 21L

L = 0.51 m

The center of gravity is 0.51 m to the right of the support.

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3 years ago

If a spacecraft that seems to be motionless is deep space is given some type of quick push, what will happen?

Sergio [31]

According to the first Newton Law of Motion (sometimes called <u><em>Law of Inertia</em></u>):

An object at rest or describing a uniform straight line motion (<u>moving at constant velocity)</u>, will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion.

In other words:

An object or body will keep its state of motion until an external force changes its state

<u>This means that </u><u>objects tend to remain in its state of motion</u><u>, and is the definition of the inertia</u>, as well.

Now, going back to the spacecraft in deep space of the question, if some force acts on it, the spacecraft will change its state and move in the same direction of the force applied.

It will also keep this path until another <u>external force</u> <em>stops</em>, <em>accelerates</em> or <em>changes</em> its direction.

Therefore, the right answer is C:

The spacecraft will move and will not stop until it is stopped by an equal and opposite force.

Options A, B and D are <u>not correc</u>t because:

A) The spacecraft cannot stop by its own; this would break Newton’s Laws of Motion

B) In deep space there is not air resistance, is <u>vacuum.</u>

D) The spacecraft works in space

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3 years ago

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What rock makes up the majority of the mares?

zzz [600]

Igneous rock makes up the majority of the mares. Because of volcanic eruption.

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4 years ago

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You wish to place a spacecraft in a circular orbit around the earth so that its orbital speed will be 4.00×103m/s. What is this

Levart [38]

Answer:

The radius of orbit= $2.49\times 10^7 m$

Explanation:

We are given that

Orbital speed= $4.00\times 10^3$ m/s

We have to find the radius of orbit of spacecraft.

We know that

Gravitational constant= $6.67\times 10^{-11}m^3/kgs^2$

Mass of earth= $5.972\times 10^{24} kg$

Orbital speed= $\sqrt{\frac{GM}{r}}$

Where G= Gravitational constant

M=Mass of earth

r=Radius of orbit

Substitute the values in the formula

$4\times 10^3=\sqrt{\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}$

Squaring on both sides

$16\times 10^6=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{r}$

$r=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{16\times 10^6}$

$r=2.49\times 10^7 m$

Hence, the radius of orbit= $2.49\times 10^7 m$

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4 years ago

If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if

umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

$P'=P+\rho gh$

Put the value into the formula

$P'=1.01\times10^{5}+1025\times9.8\times10$

$P'=201450\ Pa$

We need to calculate the volume at the surface

Using equation of ideal gas

$PV= RT$

So, for both condition

$PV=P'V'$

Put the value into the formula

$V=\dfrac{201450\times5.5}{1.01\times10^{5}}$

$V=10.97\ L$

Hence, The volume at the surface is 10.97 L.

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