What is the current when the resistance is 5 ohms and the voltage is 10 volts?

If you go to columbiana middle put your real name

ololo11 [35]

Answer:

its in my user

Explanation:

5 0

4 years ago

Read 2 more answers

A paint brush is dropped from the top of a tall ladder and free falls to the ground. What changes, if any, would be observed of

muminat

Answer:

Explanation:

As the paint brush is dropped from the top of a tall ladder it started free falling

i.e. velocity of brush increases as it falls down due to the acceleration provided by gravity.

We know acceleration due to gravity is the attraction experienced by the object due to earth attraction

Value of acceleration due to gravity is constant i.e. $9.8\ m/s^2$

Therefore the correct choice is

velocity is increasing and acceleration is constant

4 0

4 years ago

The beat your doctor listens to through a sethoscope is the sound of the four values opening

djyliett [7]

and closing .

The heart has 4 valves. They are what makes the lub-dub lub-dub sounds that can be heard from the chest.

The mitral valve is located between the left atrium and the left ventricle. It closes the left atrium to collect oxygenated blood from the lungs and opens to pass it on to the left ventricle.

The tricuspid valve is located between the right atrium and the right ventricle. It closes the right atrium to hold unoxygenated blood and opens to pass it on to the right ventricle ensuring a one way flow.

The aortic valve is located between the aorta and the left ventricle. It closes the left ventricle and opens to the aorta to pass on the oxygen-rich blood to the body.

The pulmonary valve is located between the pulmonary artery and the right ventricle. It closes off the right ventricle and opens to pass on unoxygenated blood to the lungs.

5 0

3 years ago

A truck is moving at 2.0 m/s accelerates at a rate of +2.00 m/s2. It does this over a distance of 400.0m. Find the final velocit

otez555 [7]

If v is the truck's final velocity, then

v² - (2.0 m/s)² = 2 * (2.00 m/s²) * (400.0 m)

v² = 1604 m²/s²

v = 40.05 m/s ≈ 40 m/s

3 0

3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$