All categories

3 years ago

Amount of work done by a rotating object

2 answers:

5 0

The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] * [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

riadik2000 [5.3K]3 years ago

5 0

There is no question here try to add a list of options and repost it

You might be interested in

A sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 1

hammer [34]

The specific heat of the substance will be 0.129 J/g°C.

<h3>What is specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The given data in the problem is;

Q is the amount of energy necessary to raise the temperature = 3,000.0 j

M is the mass= 0.465 kg.

Δt is the time it takes to raise the temperature.=50°c

s stands for specific heat capacity=?

Mathematically specific heat capacity is given by;

$\rm Q= MC \triangle t \\\\ C = \frac{Q}{M\triangle t} \\\\ C = \frac{3000}{0.465 \triangle 50} \\\\ C =129.0 J/Kg^0C \\\\ C= 0.129 J/g^0C$

Hence the specific heat of the substance will be 0.129 J/g°C.

To learn more about the specific heat capacity refer to the link brainly.com/question/2530523

5 0

2 years ago

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

barxatty [35]

Answer:

The mass of blood is $m= 5.7876kg$

The number of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the question we are told that

The volume of blood is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of the blood is $\rho_b = 1060 kg/m^3$

% of blood that is cell is = 45.0%

% of the blood that is plasma is = 55.0%

density of blood cell is $\rho_d = 1125kg/m^3$

% of cell that are white is = 1%

% of cell that is red is = 99%

The diameter of the red blood cell is = $7.5 \mu m = 7.5*10^{-6}m$

The radius of the red blood cell is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

Generally the mass is mathematically represented as

$m = \rho_b * V_b$

Substituting value

$m = 1060 * 0.00546$

$m= 5.7876kg$

Mass of cell is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The volume of white blood cell is $V_w$ = 1% of volume of cells

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The volume of red blood cells is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The number of red blood cell is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The Number of white blood cell is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The total number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

4 years ago

Read 2 more answers

Which of the following is a physical property?

coldgirl [10]

Flammability, I think!

6 0

3 years ago

Read 2 more answers

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

For states with larger values of n that admit a d subshell, the d subshell is at even higher energy than the p subshell. In fact

frosja888 [35]

Answer:

Explanation:

Correct order of energy of energy states within an atom is as follows

1s , 2 s, 2 p 3 s ,3 p, 4s , 3d, 4p, 4p,5s etc.

There is discrepancies in this order due to overlapping of outer orbitals with inner orbitals so sometimes outer orbitals have lower energy.

4 0

3 years ago