How much force is on a cart when the acceleration of the cart is 2.6 m/s2

During a particular thunderstorm, the electric potential difference between a cloud and the ground is vcloud - vground = 3.50 10

wariber [46]

<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>

8 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

2 years ago

If light travels from oil (slower medium) to water (faster medium) at an angle, what happens to the direction of the light ray i

iVinArrow [24]

If light travels from oil to water at an angle, what happens to the direction of the light ray in water with respect to the normal, is it moves away from the normal.

7 0

2 years ago

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2

dangina [55]

A. 441 m B: 46.0 m/s

5 0

2 years ago

How would this sentence best be punctuated? She was a success

svetlana [45]

Answer:

She was a success.

i don't think commas are necessary because it's not a compound or complicated sentence.

6 0

3 years ago