What does the variable p stand for in physics?

If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)

taurus [48]

Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20 Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

In calories, 4.2 J = 1 Calorie
= 8 400 000 / 4.2 = 200 000

Q = 200 000 Calories

4 0

3 years ago

Read each description below. Choose the force diagram (free-body diagram) that best represents the description. You may neglect

Papessa [141]

A. Diagram A
B. Diagram C & D
C. Diagram B
D. Diagram C & D
E. Diagram B
F. Diagram C & D
These are simplified representations of an object's body and the force vectors acting on it. Some of the main forces that are involve are normal force, friction, push or pull and gravity.

5 0

3 years ago

Density of water 1000kgper m^3 What will be the volume of 3500 kg water

skad [1K]

Answer:

The volume of water is 3.5 cubic meter

Explanation:

Density

The density of a substance or material is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

We are given the density of water as $\rho=1,000\ kg/m^3$ .It's required to find the volume of m=3,500 kg of water. Solving for V:

$\displaystyle V=\frac{m}{\rho}$

$\displaystyle V=\frac{3,500}{1,000}$

$V=3.5\ m^3$

The volume of water is 3.5 cubic meter

5 0

3 years ago

Which statement correctly lists the type of electromagnetic waves from highest to lowest energy? gamma rays, ultraviolet waves,

ahrayia [7]

Answer:

Thank you for posting your question on Brainly.

Explanation:

The waves from highest to lowest are :

* Gamma

* X- Rays

* UV

* Visible

* Infrared

* Microwaves

* Radio-waves

I really hope this was helpful. Have a great day : )

6 0

4 years ago

A bowling ball weighing 71.2 N (16.0 lb) is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and release

Elina [12.6K]

Answer:

a) For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

b) For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

Explanation:

For this case we have the following data given:

$W= 71.2 N$ represent the weigth for the object

$R=3.8 m$ the length of the rope or the radius

$v= 4.2 m/s$ represent the velocity of the bowling ball

Part a

For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

Part b

For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

7 0

3 years ago