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2 years ago

A 6.00-m long rope is under a tension of 600 N. Waves travel along this rope at 40.0 m/s. What is the mass of the rope?

Physics

1 answer:

MAVERICK [17]2 years ago

6 0

Answer:

2.25 kg

Explanation:

Length, L = 6 m

tension, T = 600 N

velocity, v = 40 m/s

The formula for the velocity is given by

$v=\sqrt{\frac{T}{\mu }}$

where, μ is mass per unit length

$\mu =\frac{T}{v^{2}}$

$\mu =\frac{600}{40^{2}}$

μ = 0.375

m / L = 0.375

m = 0.375 x 6 = 2.25 kg

Thus, the mass of rope is 2.25 Kg.

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Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.

creativ13 [48]

Answer: a) 456.66 s ; b) 564.3 m

Explanation: The time spend to cover any distance a constant velocity is given by:

v= distance/time so t=distance/v

The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

replacing in the above expression we have

time 1= 627 s

time2 = 297 s

The distance traveled is 564,3 m

8 0

2 years ago

Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.

Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

$Q=-P_{tot}=-(P_{2}-P_{1})$

The potential energy for any point can be calculated as:

$P=m*g*h$

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

$Q=P_{1}=m*g*h$

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

$W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW$

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2 years ago

A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material

worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

$r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m$ is the radius

$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

$d=\frac{m}{V}$

where m = 1 kg is the mass. Substituting, we find

$d=\frac{1 kg}{7.1\cdot 10^{-5} m^3}=14,160 kg/m^3$

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2 years ago

A truck pulled a car of 2,350 kg a distance of 25 meters. If the car accelerates from 3 m/s to 6 m/s, whats the average force ex

faust18 [17]

Answer:

1,269 N

Explanation:

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2 years ago