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hjlf
3 years ago
7

The rate at which work is done is also the definition of which of the following?

Physics
1 answer:
gregori [183]3 years ago
7 0
B) Power
Because Power= Work/time
You might be interested in
In a photoelectric experiment it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident
kiruha [24]

Answer:

Planck’s constant is 6.9*10^{-34} Js

work function of metal is 5.846*10^{-19}

Explanation:

The maximum kinetic energy from electron from photoelectric effect is given by

\K_{max}=eV_o where V_o  is applied voltage and e is charge on electron and substituting charge on electron by 1.6*10^{-19}  and 1.00 V for V_o  

K_{max}=1.6*10^{-19}*1=1.6*10^{-19} J

Considering that the wavelength, \lambda of light used is given as 278*10^{-9} m

Energy of light is given by

E=\frac {hc}{ \lambda}   where \lambda  is the wavelength, h is Planck’s constant and c is speed of light. Taking c for 3*10^{8} m/s  

Substituting values of wavelength and speed of light we obtain

E=\frac {h*3*10^{8}}{278*10^{-9}}=h1.07914*10^{15} J

If work function is \phi  then

E=\phi + k_{max}  hence

h1.07914*10^{15} J=\phi +1.6*10^{-19} J ------- Equation 1

Energy corresponding to wavelength of 207 nm is

E=h\frac {3*10^{8} m/s}{207*10^{-9}}=h(1.45*10^{15}}) J

Maximum kinetic energy of electrons when V_o  is 2.6 V becomes

K_{max}=(1.6*10^{-19})*2.6V=4.16*10^{-19} J

From E=\phi + k_{max}  and substituting h(1.45*10^{15}}) J  for E and 4.16*10^{-19} J  for K_{max}  we have

h(1.45*10^{15}}) J=\phi +4.16*10^{-19} J ------ Equation 2

Equation 2 minus equation 1 gives

h3.71*10^{14}=2.56*10^{-19}

h=\frac {2.56*10^{-19}}{3.71*10^{14}}\approx 6.9*10^{-34} Js

Therefore, Planck’s constant is 6.9*10^{-34} Js

Recalling equation 1 and substituting back the value of h as obtained

h1.07914*10^{15} J=\phi +1.6*10^{-19} J

(6.9*10^{-34})*(1.07914*10^{15})=\phi +1.6*10^{-19} J

\ph=(6.9*10^{-34})*(1.07914*10^{15})-(1.6*10^{-19})=5.846*10^{-19}

Therefore, work function of metal is 5.846*10^{-19}

5 0
3 years ago
Why does the weight of water pulls the central part of the surface down?
Brut [27]

Answer:

Buoyancy or Upthrust

Explanation:

This is an upward force exerted by a fluid that opposes the weight of a partially or fully immersed object.

5 0
3 years ago
What is Power can be define as
Ksju [112]
Power (physics) In physics, power is the rate of doing work, the amount of energy transferred per unit time. In the International System of Units, the unit of power is the joule per second (J/s), known as the watt in honour of James Watt, the eighteenth-century developer of the steam engine condenser.
5 0
3 years ago
Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a
seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0
4 years ago
A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
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