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Law Incorporation [45]
3 years ago
5

A train accelerates from rest and covers a distance of 600m in 20s. What is the train's speed at the end of 20s? What is the fin

al velocity of the train after 20s?
Physics
1 answer:
Allushta [10]3 years ago
6 0

Answer:

v = 60 m/s

final velocity after 20 seconds depends on how long it continues to accelerate after 20 seconds. Also we do not know the direction to complete the velocity requirements of a scalar value and direction component.

Explanation:

If we assume that the acceleration is constant

d = ½at² = ½(Δv/t)t² = ½Δvt = ½(v - 0)t = ½vt

v = 2d/t

v = 2(600)/20 = 60m/s

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What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest
lina2011 [118]

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed  =  (27 - 0)  =  27 m/s
Time for the change = 10 s

Magnitude of acceleration =  (27 m/s) / (10 s)  =  2.7 m/s²  .
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3 years ago
Explain how a current is induced in the wire?
Nastasia [14]

Answer:

If a coil of wire is placed in a changing magnetic field, a current will be induced in the wire. This current flows because something is producing an electric field that forces the charges around the wire. (It cannot be the magnetic force since the charges are not initially moving). ... that determines the induced current.

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2 years ago
A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r
andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature T_1=781 K

Reservoir 2 Temperature T_2=335 K

Let Q is the amount of heat Flows i.e. Q=1477 J

thus change in Entropy is given by \frac{\sum Q}{T}

\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}

\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}

\Delta S=\frac{\sum Q}{T}=-1.891+4.4089

\Delta S=\frac{\sum Q}{T}=2.517 J/K                              

6 0
2 years ago
An astronaut is taking a space walk near the shuttle when her safety tether breaks. what should the astronaut do to get back to
sergeinik [125]
The best way in handling in this situation is that in order for the astronaut to be able to get back to the shuttle is that he or she should take an object from his or her tool belt and to be thrown out away from the shuttle. This will allow her to weight lightly and safely return to the shuttle and would be easier for his or her to do so.
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3 years ago
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are
igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

4 0
3 years ago
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