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3 years ago

NEED HELP ASAP!!!

Physics

1 answer:

sineoko [7]3 years ago

5 0

Answer:

2,500 watts

Explanation:

I got this answer right on a test. I hope it works for you to.

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The plane of a5cm*8cm rectangular loop of wire is parallel to a 0.19T magnetic field the loop carries a current of 6.2 A. What t

Sedaia [141]

Answer:

Torque; τ = 4.712 × 10^(-3) J

Magnetic moment; M = 0.0248 J/T

Explanation:

Torque is gotten from the formula;

τ = BIA

Where;

B is magnetic field

I is current

A is area

We are given;

B = 0.19T

I = 6.2A

Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m

Thus;

Area; A = 0.05m × 0.08m = 0.004 m²

Thus;

τ = 0.19 × 6.2 × 0.004

τ = 4.712 × 10^(-3) J

Formula for the magnetic moment is given by;

M = IA

M = 6.2 × 0.004

M = 0.0248 J/T

5 0

3 years ago

How Is radar used to determine the speed of a car

Serga [27]

Radars are frequently used to identify distance and speed, such as how far away an object is or how fast it is moving. The radar device can then use the change in frequency to determine the speed at which the car is moving. In laser-speed guns, waves of light are used in place of radio waves.

4 0

3 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

2 years ago

Physics Help ASAP Please!

balu736 [363]

Answer:

23N

Explanation:

5 0

3 years ago

Read 2 more answers

Use V=Voe^-t/RC and T=RC to derive T1/2=Tln2

Ksju [112]

Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.

3 0

3 years ago