|acceleration| = (change in speed) / (time for the change)

   = (10 m/s - 0) / (4 s)

   = (10 / 4) (m/s²)

   = 2.5 m/s² .

The direction of the acceleration is west.

3 0

3 years ago

A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ

Alborosie

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, $C_1=15\ \mu F$

Capacitance, $C_2=25\ \mu F$

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

$C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F$

The energy stored in the capacitor is given by :

$E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J$

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

4 0

3 years ago

A boy is pedaling his bicycle at a velocity of 0.20km/ minute . How far will he travel in 2.5 hours

fredd [130]

30km. 24 the first two hours and 6 the half hour

4 0

4 years ago

Read 2 more answers

Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

$K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}$

Final kinetic energy is :

$K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}$

Now, fraction of initial kinetic energy loss is :

$Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25$

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0

3 years ago