What is the net force on the object represented by the FBD below?

If one doubles the emfs in a circuit and doubles the resistances in the circuit at the same time, what happens to the currents t

bija089 [108]

Answer:

Same

Explanation:

Let R be the resistance of the cell and E be the emf.

Let i be the current in the circuit.

i = E / R ..... (1)

Now emf and resistance be doubled.

so, i' = 2E/2R = E/R

So, i' = i (From equation (1)

Current remains same.

3 0

3 years ago

Homework B5

nydimaria [60]

Answer:

I hope this helps a little bit.

7 0

3 years ago

A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 23.8 s. Assuming constant angular acceleration, what is it

ser-zykov [4K]

Answer:

$0.053 rad/s^2$

Explanation:

0.2 rev/s = 0.2 rev/s * 2π rad/rev = 0.4π rad/s

Since the angular acceleration is assumed to be constant, and the wheel's angular speed is increasing from rest (0 rad/s) to 0.4π rad/s within 23.8s. Then the angular acceleration must be

$\alpha = \Delta \omega / \Delta t = \frac{0.4 \pi - 0}{23.8} = 0.053 rad/s^2$

8 0

3 years ago

A motorist drives north for 36.0 minutes at 96.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km

Over [174]

Answer:

a) $d=187.6km$

b) $v=61.5km/h$

Explanation:

First we convert our minutes to hours so we work always in the same units.

$36min=36min(\frac{1h}{60min})=0.6h$

$15min=15min(\frac{1h}{60min})=0.25h$

Where we used the fact that 1 hour are 60 min, thus the multiplying factor is equal to 1 (not altering the time, just changing the units).

a) On the first part the motorist travels a distance $d_1=v_1t_1=(96km/h)(0.6h)=57.6km$ , and on the second part he travels $d_2=130km$ .

The total displacement is $d=d_1+d_2=57.6km+130km=187.6km$

b) The average velocity is the relation between the total displacement and the time taken to cover it. Our total time is t=0.6h+0.25h+2.2h=3.05h, thus we have:

$v=\frac{187.6km}{3.05h}=61.5km/h$

8 0

3 years ago

A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s

expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

$\vec v_f=\vec v_o+\vec a.t$

$\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}$

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is $2.7 m/s^2$ to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas: