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tangare [24]
3 years ago
12

A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the

spring has a height of 82 cm and a constant of 24 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
Physics
1 answer:
krek1111 [17]3 years ago
5 0

Answer:

45.6 cm

Explanation:

Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation

E_g = E_p

mgh = 0.5kx^2

where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.

4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2

47.088x + 227.906 = 1200x^2

1200x^2 - 47.088x - 227.906 = 0

x = 0.456m or 45.6 cm

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The force F shown in Figure 4.30 has a moment of 40 Nm about the pivot. Calculate the magnitude
salantis [7]

\boxed{\sf \tau=rFsin\theta}

Put values

\\ \rm\hookrightarrow 40=2Fsin40

\\ \rm\hookrightarrow Fsin40=20

\\ \rm\hookrightarrow 0.64F=20

\\ \rm\hookrightarrow F=31.25N

8 0
2 years ago
A chunk of ice with a mass of 1.2 kg melts and absorbs 3.33 x 10³J of energy. Which best describes what has happened in this sys
Maurinko [17]

Answer:

- Its entropy increases.

Explanation:

Entropy is defined as a 'measure of the amount of energy in a physical system that cannot be used to do work.' It is also employed to denote randomness, disorder, or uncertainty of the arrangement/system. In the given system, the melting of ice denotes the 'increase in entropy' as the amount of energy unavailable to do work increases('absorbs 3.33 x 10³J of energy'). Thus, <u>this signifies that the entropy increases with a rise in temperature as it allows the substance to have greater kinetic energy</u>.

3 0
3 years ago
Convert 3402kgm/s to 20000Newtons
oee [108]

The 3,402 has units of kg-m/s.  That's momentum.  The 20,000 has units of Newtons.  That's force.  Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely.  You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push.  The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0
3 years ago
Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn
mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

5 0
3 years ago
A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors
ale4655 [162]

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, F_1 = 140 N

let the second force = F_2

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N

F_2 = 280 \ East

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N

F_2 = 560 \ West

8 0
3 years ago
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