What is a force that slows things down and uses up energy?

if the mass of the objects stays the same and their distance from eachother decreases, then the force of gravity between the two

dedylja [7]

Then the force of gravity between them would be quadrupled and so on but the gravitational force is inversely proportional to the square of the separation distance between the two interacting objects which makes more separation distance will result in weaker gravitational forces

5 0

1 year ago

According to the inverse square law of light, how will the apparent brightness of an object change if its distance to us triples

aleksley [76]

According to the inverse square law of light, <span>apparent brightness will decrease by a factor of 9. Use the formula </span> $B=L/(4*pD^2)$ , to check it.

4 0

3 years ago

A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant

Alla [95]

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v $_f$ + xv $_{fg$

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v $_f$ + x₂v $_{fg$

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

3 0

3 years ago

A rocket, blasting into space, accelerates at 2 m/s^2 for 20 seconds. What is its speed at the end of 20 seconds?

OLEGan [10]

Answer:

Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (beginning speed)

= (3,600 mi/hr) - ( 0 )

= 3,600 mi/hr

= same as 1 mile/second.

Acceleration = (1 mi/sec) / (10 sec)

= 0.1 mi/sec² .

But 1 mile = 5,280 ft,

so

0.1 mi/s² = 528 ft/s²

Explanation:

have a great day

4 0

3 years ago

A dog pulls on a pillow with a force of 5 N at an angle of 37o to the horizontal. Find the x and y components of this force.

Advocard [28]

Answer:

x- component = Fx= 4N

y - component= Fy= 3N

Explanation:

Fx=Fcos∅= 5cos 37= 3.99≅4N

Fy=F sin∅= 5 Sin37= 3.009≅3.01N

the dog pulls the pillow at an angle 37 degrees to the horizontal means it pulls it at an angle 37 degrees with the x- axis.if the line of the force represents the vector F then its horizonatal and vertical components wil be along x-axis and y-axis ie Fx and Fy.

as sin∅= Perpendicular/ Hypotenuse

but we have force F and Fy

so sin∅= Fy/F

or Fy= Fsin∅---------eqA

similarly cos∅= base /hpotenuse

cos∅= Fx/F

Fx= Fcos∅----------eq B

eq A and eq B can be used to determine the values of the rectangular components of the force making an angle with the horizontal

5 0

3 years ago