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2 years ago

Find the current through a 12 ohm resistive circuit when 24 volts is applied

Physics

1 answer:

marusya05 [52]2 years ago

5 0

Answer:

Explanation:

Voltage = Current * Resistance

V= I*R

I=V/R

I = 24/12 = 2A.

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HELPPP PLEASE !!!!!!!

vlabodo [156]

Explanation:

a) d = ½.a.t²

200 = ½(4)t²

200 = 2t²

t² = 200/2

t² = 100

t =√100 = 10 s

b) Vt = a. t

= 4(10)

= 40 m/s

c) V av. = d/t = 200/10 = 20m/s

6 0

1 year ago

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

Roman55 [17]

Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

8 0

2 years ago

Read 2 more answers

Help please <br> hhjshwjsjejjenrhrhfhhfheisiw

DanielleElmas [232]

Answer:

Explanation:

4 0

2 years ago

Sheila uses a 45N force on her bowling ball across a 15m lane. What work did she do on the bowling ball? Show your work.

Maurinko [17]

Answer:

675J

Explanation:

Given parameters:

Force = 45N

Distance = 15m

Unknown:

Work done by Sheila = ?

Solution:

Work done by a body is the amount of force applied to make a body move through a distance;

Work done = Force x distance

Now;

Work done = 45 x 15 = 675J

6 0

2 years ago

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

2 years ago