The horizontal force applied is 160 N while the velocity is 2.03 m/s.
<h3>What is the speed of the car?</h3>
The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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Answer: 2. Solution A attains a higher temperature.
Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.
In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.
Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.
<em>We have a formula for such condition,</em>
.....................................(1)
where:
= temperature difference
- c= specific heat of the body
<u>Proving mathematically:</u>
<em>According to the given conditions</em>
- we have equal masses of two solutions A & B, i.e.

- equal heat is supplied to both the solutions, i.e.

- specific heat of solution A,

- specific heat of solution B,

&
are the change in temperatures of the respective solutions.
Now, putting the above values


Which proves that solution A attains a higher temperature than solution B.
Answer:
Therefore, we need an invert, and a rectifier, along with the transformer to do the job.
Explanation:
A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.
Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.
So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.
<u>Therefore, we need an invert, and a rectifier, along with the transformer to do the job.</u>
Pretty sure it’s false....................