All categories

3 years ago

Two resistors, R1 and R2, are connected

Physics

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

18 ohms

Explanation:

V = I(R1 + R2)

5V = (0.167A)(12 ohms + R2)

Solving for R2

R2 = 18 ohms

You might be interested in

A 10 µf capacitor is charged to 108 v and is then connected across a 328 ω resistor. what is the initial charge on the capacitor

valina [46]

The capacitance is defined as the maximum charge stored in a capacitor, Q, divided by the voltage applied, V:
$C= \frac{Q}{V}$

The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
$Q=CV=(10 \mu F)(108 V)=1080 \mu C$

8 0

4 years ago

A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at

kherson [118]

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity u and it falls under the acceleration due to gravity g.

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

$s=ut+\frac{1}{2}at^2$

The bag makes a net displacement s of 13.4 m downwards, hence

$s=-13.4 m$

Its initial velocity is

$u=+4.6 m/s$

The acceleration due to gravity acts downwards and hence it is negative.

$g=-9.8 m/s^2$

Use the values in the equation of motion and write an equation for t.

$s=ut+\frac{1}{2} at^2 \\ -13.4=4.6t-\frac{1}{2}(9.8)t^2\\ 4.9t^2-4.6t-13.4=0$

Solving the equation for t and taking only the positive value for t,

t=2.18 s

4 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$