Radar, Radio, Microwaves are examples of this kind of energy im so confused on what energy this is

PLEASE HELP!!!!!!!!! 30 POINTS

ivolga24 [154]

Answer: experiment data is the things you do in the experiment and the result is the answer

4 0

3 years ago

How many main retrofit strategies help solve the ecological problems with suburbs?

iragen [17]

ANSWER:

5

Explanation:

Because they are elven in numbers

3 0

2 years ago

An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef

Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field $B=5.0\times10^{-5}\ T$

(a). We need to calculate the induced emf

Using formula of emf

$E = Blv\sin\theta$

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

$E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}$

$E=1.398\times10^{-3}\ V$

We need to calculate the induced current

$E =IR$

$I=\dfrac{E}{R}$

Put the value into the formula

$I=\dfrac{1.398\times10^{-3}}{3.0}$

$I=0.5\ mA$

(b). We need to calculate the power dissipated through the resistor

Using formula of power

$P=I^2 R$

Put the value into the formula

$P=(0.5\times10^{-3})^2\times3.0$

$P=7.5\times10^{-7}\ Watt$

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and $7.5\times10^{-7}\ Watt$ .

6 0

3 years ago

Read 2 more answers

Which one has greater potential energy? An 8 kg rock sitting on a 2.2 m cliff or a 6 kg rock sitting on a 3.2 m cliff.

Evgesh-ka [11]

Answer: The 6 kg rock sitting on a 3.2 m cliff.

Explanation:

The potential energy of an object of mass M that is at a height H above the ground us:

U = M*H*g

where g is the gravitational acceleration:

g = 9.8m/s^2

Then:

"An 8 kg rock sitting on a 2.2 m cliff"

M = 8kg

H = 2.2m

U = 8kg*2.2m*9.8 m/s^2 = 172.48 J

"a 6 kg rock sitting on a 3.2 m cliff"

M = 6kg

H = 3.2m

U = 6kg*3.2m*9.8m/s^2 = 188.16 J

You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.

6 0

3 years ago

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located

andreev551 [17]

Answer:

The correct answer is 32.9 m/s

Explanation:

To solve this, we list out the known and the unknown variables as follows

Maximum allowable acceleration = 1.34 m/s²

Distance between sttions = 806 m

Therefore from the equation of motion

v = ut + 0.5·×at²

Where v = final velocity

u = initial velocity

S = distance covered

t = time

a = acceleration

Also v² = u² + 2·a·S

where u is the initial velocity, which we can take as u = 0, then

v² = 2·1.34·S = 2.68S m²/s² then

Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z, so u² = 2.68Z

since u² = 2.68S from the previous calculation, then for v = 0

2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m

and v² = 2.68S m²/s² = 1080.04 m²/s²

v = 32.9 m/s

The maximum speed a subway train can attain between stations is 32.9 m/s

3 0

3 years ago