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ikadub [295]
3 years ago
15

On a part time job you are asked to bring a cylindrical iron rod of density 7800 kg/m^3, length 81.2 cm and diameter 2.75 cm fro

m a storage room to a machinist. Calculate the weight of the rod, w. The acceleration due to gravity, g= 9.81 m/s^2.
w= 36.9 N Part A

Part B) Now that you know the weight of the rod, do you think that you will be able to carry the rod without a cart? Yes or No?
Physics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

3.76188 kg

36.9040428 N

Yes

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

L = Length of rod = 81.2 cm

d = Diameter of rod = 2.75 cm

r = Radius = \frac{d}{2}=\frac{2.75}{2}=1.375\ cm

A = Area of the cylinder =\pi r^2

v = Volume = L\times A

m = Mass

\rho = Density of rod = 7800 kg/m³

Density

\rho=\dfrac{m}{v}\\\Rightarrow m=\rho\times v\\\Rightarrow m=\rho\times L\times A\\\Rightarrow m=7800\times 0.812\times \pi\times 0.01375^2\\\Rightarrow m=3.76188\ kg

The mass of the rod is 3.76188 kg.

As the the mass of the rod is 3.76188 kg I will be able to carry the rod.

Weight

W=mg\\\Rightarrow W=3.76188\times 9.81\\\Rightarrow W=36.9040428\ N

The weight of the rod is 36.9040428 N

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3 years ago
A bowling ball has a mass of 5 kg. What happens to its momentum when its speed increases from 1 m/s to 2 m/s?
faltersainse [42]

The linear momentum is given by mv. Here m is mass v is velocity of the body. Given mass and velocities as m=5,v_i=1,v_f=2

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The correct answer is (A)

8 0
4 years ago
What is the force that can cause two pieces of iron to attract each other?
Rudiy27

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B. Magnetic Force

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6 0
3 years ago
Read 2 more answers
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
ziro4ka [17]

Answer:

25 m/s

Explanation:

from the question you van see that some detail is missing, however i found this same question using internet search engines on: 'https://www.chegg.com/homework-help/questions-and-answers/light-rail-passenger-trains-provide-transportation-within-cities-speed-slow-nearly-constan-q5808369'

here is the complete question:

'Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0m/s . Once free of this area, it speeds up to 12m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final Speed?'

SOLUTION

initial speed (u) = 7 m/s

final speed (v) = 13 m/s

initial acceleration time (t1) = 8 s

final acceleration time (t2) = 16 s

what is the higher cruising speed?

acceleration = \frac{final speed (v) - initial speed(u)}{time (t1)}

acceleration = \frac{13-7}{8} = 0.75 m/s^{2}

since the train accelerates at the same rate, the increase in  speed will be = acceleration x time (t2)

= 0.75 x 16 = 12 m/s

therefore the higher cruising speed = increase in speed + initial speed

= 12 + 13 = 25 m/s

4 0
3 years ago
What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?
Marta_Voda [28]

Answer:

The answer is \frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709. Let's learn why.

Explanation:

Newton's law of universal gravitation says;

F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}

Here G is a universal gravitational <u>constant</u> and is measured experimentally.

Sun's gravitational pull on mercury is:

F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }

Therefore F_{Sun-Mercury} = Gm_{sun} 98,4366

Sun's gravitational pull on Earth is:

F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}

Therefore F_{Sun-Earth} =Gm_{sun} 265,33

As a result;

\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709

4 0
4 years ago
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