All categories

3 years ago

If we increase the force applied to an object, and all other factors remain the same, the amount of work will

Physics

2 answers:

Vesnalui [34]3 years ago

6 0

Answer:

C.Increase

Explanation:

Let F be the force applied on the object.

Displacement of object=x

Now, work done by the objects= $Force\times displacement$

Apply the formula

Then , we get

Work don=w= $F\times x=F\cdot x$

When we increase the force applied on the object and all other factors remain same means displacement remain same.Then what effect on work.

We know that work done is directly proportional to force and displacement.

If we increase the force then work done is also increase.

Answer:C.Increase

soldier1979 [14.2K]3 years ago

3 0

The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.

-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.

-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.

-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.

You might be interested in

Who is the scientist that gave us the photon as a way of describing light as a particle?

Montano1993 [528]

Answer: Albert Einstein

Explanation:

Light can be considered as a wave or as particles, in this context Einstein proposed that light behaves like a stream of particles called <u>photons</u> with an energy, in order to correctly explain the photoelectric effect (in fact he won the 1921 Nobel Prize in Physics because of this explanation).

To uderstand it better:

The photoelectric effect is a fenomenom that consists in the emission of electrons that occurs when light falls on a metal surface under certain conditions.

This can only be explained based on the corpuscular model of light, that is, light is quantized.

So, Einstein theorized light as a stream of energy packets called photons, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

4 0

3 years ago

Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a

11111nata11111 [884]

Answer:

The box 1 moves faster.

Explanation:

lets

Mass =m kg

Initial velocity = u m/s

Initial velocity of box = 0 m/s

Let stake mass of block = m

When ball bounces back:

The final speed of the box = v

Final speed of ball = - u

Pi = Pf ( From linear momentum conservation)

m x u + m x 0 = m ( - u) + m v

mu + mu = m v

v= 2 u

When ball get stuck :

The final speed of ball and box = v

Pi = Pf ( From linear momentum conservation)

m x u + m x 0 = (m+m) v

v= u /2

So the box 1 moves faster.

6 0

4 years ago

ayo 20 m long is heated from a temperature of 5 degree celsius to 55 degree celsius if the change in length is 0.020 m calculate

Lera25 [3.4K]

Answer:2 x 10^-5K^-1

Explanation:

The solution is in the attached file

6 0

2 years ago

~ is the following statement true or false? Explain your answer.

USPshnik [31]

Answer: False

Explanation: The sun is one of earths primary energy sources. Without the sun, all animals, plants, humans would die. The sun's energy provides warmth for humans and plants and animals cannot grow without the sun.

7 0

3 years ago

Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure, $P_1=14.7psia,T_1=60\textdegree F$

#Compressor outlet:

$Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min$

#Cooling rate,

$q_{out}=10Btu/lbm, \dot W=700hp$

# From table A-1E

Gas constant of air $R=0.3704\ psia.ft^3/lbm.R$

Specific enthalpy at $P_1=520R-h_1=124.27Btu/lbm$

Using the mass balance:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F$

Hence, the temperature at the compressor exit $T_2=341\textdegree F$

5 0

4 years ago