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shusha [124]
3 years ago
8

Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica

l burners. Afterward, liquid A is hotter than liquid B. Which has the larger specific heat?
Physics
1 answer:
DochEvi [55]3 years ago
5 0

Answer:

So the specific heat of the liquid B is greater than that of A.

Explanation:

Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

Q=m.c.\Delta T

where:

m = mass of the body

c = specific heat of the body

\Delta T= change in temperature of the body

The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.

So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.

\Delta T=\frac{Q}{m} \times \frac{1}{c}

\Delta T\propto\frac{1}{c}

So the specific heat of the liquid B is greater than that of A.

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The time period T of a simple pendulum is given by the relation
Vanyuwa [196]

Answer:

T^2 \propto L

Explanation:

The period of a simple pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

From this equation we can write

T\propto \sqrt{L}\\T\propto \frac{1}{\sqrt{g}}

Taking the square of this equation, we get:

T^2 = (2\pi)^2 \frac{L}{g}

So we see that T^2 is proportional to L and inversely proportional to g. So, we can write:

T^2 \propto L\\T^2 \propto \frac{1}{g}

So the only correct option is

T^2 \propto L

5 0
4 years ago
Read 2 more answers
A cyclist is taking part in the Tour de France, which is a bicycle race that takes place every year.
marishachu [46]
Solve the following word problems.
1. The ratio of red marbles and blue marbles that Carlo has is 8: 3. When he
exchanged 35 red marbles for 20 blue marbles from his brother, he was left with
equal number of red and blue marbles.
How many red and blue marbles did he have at the beginning
How many red and blue marbles did he have now
5 0
3 years ago
At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta
aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given  :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.  

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next  

F=|F|cosФp            (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet  

F=|F|cosФp

 =-4.8*10^22 N*p

<em>As this force is in one direction, we could get its vector as next  </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0
3 years ago
Your go-cart breaks down right before the end of a race, so you have to push it over the finish line. The go-cart has a mass of
Flura [38]

a) W=mg=833 N by definition

b) F=ma=76.5 N according to Newton's second law

c) a=F/m=4.12 m/s^2

d) m=F/a=720/5.5=130.91 kg

6 0
3 years ago
Calculate the kinetic energy of an 800 kg car when it is going at
Mashcka [7]

Answer:

I A 1000 kg car is traveling at 20 m/s. How much kinetic energy does it. Here, Given- mass=1000kg Speed=20m/s Formula ...

Missing: 13.4m/s ‎| Must include: 13.4m/s

Explanation:

3 0
3 years ago
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