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3 years ago

HURRY!!! I need helppppp!!!

Physics

1 answer:

RoseWind [281]3 years ago

5 0

Answer:

ok i think the answer would be C. or B. hope im right

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A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle

Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

$\tau=NIABsin\theta$ (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

$B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T$

3 0

3 years ago

A piece of taffy slams into and sticks to another identical piece of taffy that is at rest. The momentum of the two pieces stuck

-Dominant- [34]

Answer:

C. 50%

Explanation:

Lets consider that the mass of taffy is m and its initial velocity is u and final velocity is v. Momentum is conserved so we can write it as,

$mu = 2mv\\v = \frac{u}{2} \\$

The initial kinetic energy = $\frac{mu^{2} }{2}$

As the Kinetic energy partly converted into heat.

The final kinetic energy = $\frac{m(\frac{u}{2})^{2} }{2}$ = $\frac{mu^{2} }{4}$

Change in the kinetic energy = $\frac{mu^{2} }{2} - \frac{mu^{2} }{4} = \frac{mu^{2} }{4}$

now we can determine the fraction of kinetic energy that has turned into heat

= $\frac{Heat energy}{initial K.E}$ = $\frac{\frac{mu^{2} }{4} }{\frac{mu^{2} }{2} } = 1/2$

= 50%

3 0

4 years ago

A blank is used to control the temperature in a room, please help

kow [346]

Thermostat is used to control the temperature in a room

3 0

3 years ago

Read 2 more answers

A weather balloon is designed to expand to a maximum radius of 21 m at its working altitude, where the air pressure is 0.030 atm

dezoksy [38]

Answer:

7.65 m

Explanation:

$P_1$ = Initial pressure = 0.03 atm

$P_2$ = Final pressure = 1 atm

$r_1$ = Inital radius = 21 m

$V_1$ = Intial volume of gas = $\frac{4}{3}\pi r_1^3$

$V_2$ = Final volume of gas = $\frac{4}{3}\pi r_2^3$

$T_1$ = Initial temperature = 200 K

$T_2$ = Final temperature = 323 K

From ideal gas law we have

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m$

The radius at liftoff is 7.65 m

7 0

4 years ago

At a given instant in time, a 6-kg rock that has been dropped from a high cliff experiences a force of air resistance of 15 n. w

Vinil7 [7]

<span>Gravitational force is equal to 6 kg * 9.81 m/s^2 = 58.86 N. If air resistance is 15 n up, the magniture is 43.86 N downward.</span>

5 0

3 years ago