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BigorU [14]
3 years ago
15

A student pushes a box across a rough horizontal floor. If the amount of work done by the student on the box is 100 J and the am

ount of energy dissipated by friction is 40 J, what is the change in kinetic energy of the box
Physics
1 answer:
Archy [21]3 years ago
4 0

Answer:

60J

Explanation:

friction is the opposition to the motion of an object. Before a body can be set in motion, part of the energy or work-done must be used to overcome friction.

While kinetic energy is the energy associated with a body due to its motion.

Since friction tends to oppose the motion of a body,when a body start to move, some of the energy applied to the body to cause its motion is converted to heat due to friction.

Hence if a work-done of 100J is applied to a body that requires 40J of the work to overcome friction, the total change in the kinetic energy can be expressed as

K.E=100J-40J\\K.E=60J\\

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The milky way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to hubble
lora16 [44]
SBb type spiral. Its a type B because its not too tightly wound but its still too tight to be a type C
4 0
3 years ago
Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in
iVinArrow [24]

Answer:

The value is  P_2 = 40.54 \ psla

Explanation:

From the question we are told that

 The initial pressure is  P_1 = 14\  psla

  The initial temperature is  T_1 =  50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283  \  K

   The final temperature is  T_2 =  320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433  \  K

Generally the equation for adiabatic process is mathematically represented as

         PT^{\frac{\gamma}{1- \gamma} } =  Constant

=>      P_1T_1^{\frac{\gamma}{1- \gamma} } =  P_2T_2^{\frac{\gamma}{1- \gamma} }

Generally for a monoatomic gas  \gamma =  \frac{5}{3}

So

           14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }

=>       14 * 283^{-2.5} =P_2 * 433^{-2.5}

=>       P_2 = 40.54 \ psla

8 0
2 years ago
1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

8 0
3 years ago
Does mars has a bulge near its equator ?
Keith_Richards [23]
Yessir it sure does
7 0
3 years ago
We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e
Tju [1.3M]

The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>
  • So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
  • These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
  • Since the Moon and Sun are on opposite sides of Earth during a full moon, the light's distance traveled equals,

         d=(1.5*10^8km)+2(3.8*10^5km)=1.51*10^8km=1.51*10^{11}m

  • As we know that light travels at a speed of 300,000 km per second. then, the time taken by the light reflected from sun to reach on earth will be,

                      t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

Learn more about distance here:

brainly.com/question/11495758

#SPJ4

6 0
2 years ago
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