PLS HELP ITS URGENT HELP PLSSS

A 380-N girl walks down a flight of stairs so that she is 2.5 m below her starting level. What is the change in the girl's gravi

Svetach [21]

PEg = Mass x Height x Gravity. So it doesn't matter how much PEg she started with, since you're finding how much it changed when she walked down 2.5 meters of stairs. So by plugging into the equation, you'll find how much potential energy was lost by walking down. (I'll leave it up to you since idk if you use 10 or 9.8 as gravity)

8 0

3 years ago

Read 2 more answers

A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o

Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: 

Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is:

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.

Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)

W = ∫ 1250π x dx
W = 1250π ∫ x dx
W = 625π * (x2 – x1)

x2 = 14 ft + 15 ft = 29 ft

x1 = 14 ft + 1 ft = 15 ft

W = 625π * (29^2 - 15^2)
W = 385,000π ft-lbs = 1,209,513.17 ft-lbs

3 0

3 years ago

Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when s

worty [1.4K]

Answer:

a. about 20 years younger

b. Malcolm sits around for 49.94 years

c. 2.268x $10^{17}$  m

Explanation:

light travels 3x $10^{8}$ m in one seconds

in 20 years that will be 3x $10^{8}$ x 20 x 60 x 60 x 24 x 365 = 1.89x $10^{17}$ m

for the to and fro journey, total distance covered will be 2 x 1.89x $10^{17}$ = 3.78x $10^{17}$ m

Flo's speed = 80% of speed of light = 0.8 x 3x $10^{8}$ = 2.4x $10^{8}$ m/s

time that will pass for Malcolm will be distance/speed = 3.78x $10^{17}$ /2.4x $10^{8}$

= 1575000000 s = 49.94 years

the relativistic time t' will be

t' = t x $\sqrt{1 - \frac{v^{2} }{c^{2} } }$

t' = 49.94 x $\sqrt{1 - 0.8^{2} }$

t' = 49.94 x 0.6 = 29.96 years this is the time that has passed for Flo

this means that Flo will be about 20 years younger than Malcolm when she returns

relativistic distance is

d' = d x $\sqrt{1 - \frac{v^{2} }{c^{2} } }$

d' = 3.78x $10^{17}$ x $\sqrt{1 - 0.8^{2} }$

d' = 3.78x $10^{17}$ x 0.6

d' = 2.268x $10^{17}$  m this is how far it is to Flo

6 0

3 years ago

What is the correct free body diagram for the cup?

maksim [4K]

Where is the body diagram?

4 0

3 years ago

A 0.42 kg mass is attached to a light spring with a force constant of 34.9 N/m and set into oscillation on a horizontal friction

Whitepunk [10]

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$ (1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$ (2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{(34.9 N/m)(0.05 m)^2}{0.42 kg}}=0.456 m/s$

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

$E=\frac{1}{2}kA^2=\frac{1}{2}(34.9 N/m)(0.05 m)^2=0.044 J$

So, solving (3) for v, we find the speed when x=1.5 cm:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

$v=\frac{v_{max}}{2}=\frac{0.456 m/s}{2}=0.228 m/s$

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\x=\sqrt{\frac{2E-mv^2}{k}}=\sqrt{\frac{2(0.044 J)-(0.42 kg)(0.228 m/s)^2}{34.9 N/m}}=0.044 m=4.4 cm$

4 0

3 years ago