Proof that the kinetic energy of a moving body is half mv square

What does the electric field strength tell about the electric firld?

Solnce55 [7]

Answer:

Explanation:

The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.

The strength of electric field is defined as the force experienced by the unit positive test charge.

E = F / q

Electric field strength is a vector quantity and it is measured in newton per coulomb.

Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.

The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.

8 0

3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

so

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

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

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

3 years ago

A concave mirror has a radius of curvature of 0.60 meter.

love history [14]

Pls give me brainliest!!!

6 0

3 years ago

What is the maximum of the sinusoidal function?.

professor190 [17]

The max is the largest it could get so ( ,0)

8 0

2 years ago

In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t

Bumek [7]

Explanation:

Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.

Hence, $\Delta E$ = 0. And, as $\Delta E$ = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.

Therefore,

Q = -(19000 J + 2000 J)

= -21000 J

or, |Q| = 21000 J

Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.

5 0

3 years ago

Proof that the kinetic energy of a moving body is half mv square​

Proof that the kinetic energy of a moving body is half mv square