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Svetradugi [14.3K]
3 years ago
5

Suppose a train engine is pulling ten cars. The last car becomes separated from the train. What happens to the motion of the res

t of the train and why?
Physics
2 answers:
melisa1 [442]3 years ago
6 0
It becomes faster because there is less weight being pulled

scZoUnD [109]3 years ago
3 0
It will go faster because the train is pulling less weight
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Which of the following is not a subatomic particle?
Fynjy0 [20]
D. Nucleus because it is not a part of the group.
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Where do you feel that you are traveling at the fastest speed when on the swing?
il63 [147K]

Answer:

C

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

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3 years ago
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An unknown substance has a mass of 0.125 kg and an initial temperature of 95.0°C. The substance is then dropped into a calorimet
mr Goodwill [35]

Answer:

c = 1163.34 J/kg.°C

Explanation:

Specific heat capacity:

"Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass. The specific heat capacity of a material is a physical property."

Use this equation:

mcΔT = ( mw c + mAl cAl )  ΔT'

Rearranging the equation to find the specific heat (c) you get this:

c = (( mw c + mAl cAl )  ΔT') / (mΔT)

c = (( 0.285 (4186) + (0.15)(900)) (32 -25.1)) / ((0.125) (95 - 32))

c = 1163.34 J/kg.°C

7 0
3 years ago
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Select the correct answer. Brian is repairing an old alarm clock. He needs to replace a device that converts the electric energy
Stella [2.4K]

it shouldbe a buzzer

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3 years ago
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I got part c right but idk why the other parts are wrong HELP!
dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

J=\int F\Delta t

where

F is the magnitude of the force exerted on the object

\Delta t is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

B=0.14-0 = 0.14s\\b=8-5=3s

and whose height is

h=900 N

Therefore, the area (and the impulse) is

J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

J=F_{avg} \Delta t

where

F_{avg} is the average force exerted during the whole time \Delta t

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

\Delta t = 0.14 s is the time interval

Solving for the average force, we find

\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

J=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
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