A body of mass 120 g is taken vertically upwards to reach the height of 5m,claculate work done

If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0

3 years ago

What other places in the world use hydro kinetic energy to generate electrical power?

Elina [12.6K]

Answer:

Alaska: Hydrokinetic Energy Campbell CR9000X used for in-stream hydrokinetic device evaluation. Marine hydrokinetic energy power generation is an emerging sector in the renewable energy portfolio. Hydrokinetic devices convert the energy of waves, tidal currents, ocean currents or river currents into electrical power.

3 0

3 years ago

The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving

nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

$F_f = \mu_s (mg)$

where

$\mu_s = 0.5645$ is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

$F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N$

So, the applied force must be greater than this value.

3 0

2 years ago

the wattage ratting of a light bulb is the power it sonsumers when it is connected accross a 120 V potential difference what is

igor_vitrenko [27]

Answer:

240 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).

The expression for resistance power and voltage is give as,

P = V²/R.......................... Equation 1

Where P = Power, V = Voltage, R = Resistance

Making R the subject of the equation,

R = V²/P.................... Equation 2

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

Hence the resistance of the bulb = 240 Ω

8 0

2 years ago

Question

levacccp [35]

I uploaded the answer t $^{}$ o a file hosting. Here's link:

bit. $^{}$ ly/3tZxaCQ

5 0

2 years ago