Ask question

Ask question

All categories

3 years ago

14

What's the basic unit of the measurement of power

1 answer:

Jobisdone [24]3 years ago

7 0

Power is the RATE at which energy changes or flows.

The unit of energy is the Joule, so a unit of power is the Joule/second.

That unit has the special name "<em>Watt</em>".

Your question is not necessarily a question. It's also a statement, You could write

"<em>Watt's the basic unit of the measurement of power</em>".

You might be interested in

When a physical change occurs, the mass of the substance is conserved. This means that the total mass of the substance remains t

Doss [256]

Yes yes ! How right you are ! Truer words were never spoken.

7 0

3 years ago

What can be correctly concluded from this experiment?

solong [7]

1

Step by step explanation: 1

Brainlist?????

5 0

2 years ago

Read 2 more answers

Two moles of helium gas initially at 438 K and 0.44 atm are compressed isothermally to 1.61 atm. Find the final volume of the ga

docker41 [41]

Answer:

44.64335 L

Explanation:

R = Gas constant = 8.314 J/mol K = 0.08205 L atm/mol K

P = Pressure

V = Volume

T = Temperature = 438 K

1 denotes initial

2 denotes final

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2\times 0.08205\times 438}{0.44}\\\Rightarrow V=163.35409\ L$

So,

$P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{0.44\times 163.35409}{1.61}\\\Rightarrow V_2=44.64335\ L$

The volume of Helium is 44.64335 L

5 0

3 years ago

A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m

Feliz [49]

Explanation:

Let the speeds of father and son are $v_f\ and\ v_s$ . The kinetic energies of father and son are $K_f\ and\ K_s$ . The mass of father and son are $m_f\ and\ m_s$

(a) According to given conditions, $K_f=\dfrac{1}{3}K_s$

And $m_s=\dfrac{1}{4}m_f$

Kinetic energy of father is given by :

$K_f=\dfrac{1}{2}m_fv_f^2$ .............(1)

Kinetic energy of son is given by :

$K_s=\dfrac{1}{2}m_sv_s^2$ ...........(2)

From equation (1), (2) we get :

$\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}$ ..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

$\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}$

$v_s^2=4(v_f+1.5)^2$

Using equation (3) in above equation, we get :

$v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s$

(b) Put the value of $v_f$ in equation (3) as :

$v_s=7.09\ m/s$

Hence, this is the required solution.

8 0

3 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago