Please help<br>please help

A flashlight is powered by 4, 1.5-volt batteries, which are connected in series. The bulb draws 720mA or 0.72 amps when turned o

Varvara68 [4.7K]

R=0.12
Use the equation V=I/R
V= voltage (volts)
I= current (amps)
R= resistance (ohms)

3 0

4 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

3 years ago

Many gates at railway crossings are operated manually. A typical gate consists of a rod usually made of iron, consisting heavy w

alekssr [168]

Answer:

Making the crossing gate balanced makes the mechanical drive simpler and less expensive to operate.

Explanation:

See the attachment for a picture of what we're talking about.

The gate must be rotated up and down. That rotation is easier if the mass being rotated is balanced about the center of rotation. Less torque (and work) is required to create the desired motion.

7 0

3 years ago

A cat lifting a barbell of 2.5 kg a distance of 3 meters. How much work is done lifting the barbell? How much work is done if th

Degger [83]

Work = (force) x (distance)

Force = weight of the barbell = (mass) x (gravity) = (2.5 kg) x (9.8 m/s²)

Work = (2.5 x 9.8 newtons) x (3 meters) = 74 newton-meters = 74 joules.

It takes 74 joules to lift the 2.5-kg load 3 meters up off the floor.

After that, as long as the load is held motionless, no more work is done.

7 0

3 years ago

If a ball is dropped into a very viscous liquid it will

Dafna1 [17]

It will go to the bottom of the liquid very slowly. Good Luck!

7 0

3 years ago

Please helpplease help