In an experiment, it took 300s to increase the temperature of 0.1kg of the liquid from 25°C to 50°C. During this period, the ene
1 answer:
Answer:
<h2>4000Explanation:
The temperature of 0.1 kg of liquid rises from 25°C to 50°C in 300 sec. Energy of 13,600 J was supplied during this time. Appartus was losing energy at the rate of 12 J/sec.
Let us assume the Specific heat capacity as .
As there is no state change from liquid to gas, only Specific heat capacity is involved. Also, work done is approximately zero because volume does not change much. So,
Energy gained = Energy required to rise the temperature
Energy gained by liquid =
∴ Specific heat capacity of liquid = 4000
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Explanation :
It is given that an oscillator makes 360 oscillations in 3 minutes.
(a) Using unitary method :
No. of vibrations in one minute is,
So, no of vibrations in one minute is 120.
(b) Similarly,
3 minutes = 180 seconds
No of vibrations in one second is
So, the no of vibrations in one second is 2.
(c) Time period of the wave is given by :
The time period of the wave is 0.5 s
(d) The no of vibation per second is called as its frequency.
Hence, this is the required solution.
Answer:
<em>The magnitude of the pulling force is 20.66% of the gravitational force acting on the box</em>
Explanation:
<u>Accelerated Motion</u>
The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.
Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by
Solving for a
Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.
The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:
There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box
The decomposition of T gives us
Solving the above equation for N
Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:
Recalling that
Replacing N from [1]
Operating
Solving for T
We don't know the value of m, thus we'll plug in the rest of the data
Dividing by the weight of the box m.g
Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box