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AleksandrR [38]
3 years ago
13

A charge of 4.5x10^-5 C is placed in an electric field with a strength of 2.0x10^4 N/C. What is the electric force acting on the

charge?
Physics
1 answer:
Alex3 years ago
6 0

Answer:

0.9 N

Explanation:

The electric force acting on a charge is given by:

F=qE

where

q is the magnitude of the charge

E is the strength of the electric field

In this problem, we have

q=4.5\cdot 10^{-5}C is the charge

E=2.0\cdot 10^4 N/C is the strength of the electric field

Substituting into the equation, we find

F=(4.5\cdot 10^{-5}C)(2.0\cdot 10^4 N/C)=0.9 N

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PLEASE HELP ME WITH THIS ONE QUESTION
sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But,  the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0
2 years ago
A boy threw a ball upward and it took 3 sec to reach the highest point. Determine the initial velocity of the ball.
makkiz [27]

Answer:

V initial = 29.4 m.s²

Explanation:

( Using the laws of motion)

V final = V initial + Acceleration × time

0 = V initial + ( -9.8)(3)

29.4 = V initial

* I took upward as positive that's why I substituted -9.8 *

* for V final we know that at maximum height the ball is not moving thats why is = 0 *

6 0
1 year ago
The pressure P (in kilopascals), volume V (in liters), and temperature T (in kelvins) of a mole of an ideal gas are related by t
stich3 [128]

Answer:

\dfrac{dV}{dt} = -0.466 m^3/s

Explanation:

given,

P (in kilo pascals),         volume V (in liters),        temperature T (in kelvins)

P V = 8.31 T

Rate of increase the temperature =  0.1 K/s

temperature = 285 K

Pressure = 18 kPa

increasing at the rate of 0.07 k Pa/s

Rate at which volume is changing = ?

V = 8.31 \dfrac{T}{P}

\dfrac{dV}{dt} = 8.31 \dfrac{P\dfrac{dT}{dt}-T\dfrac{dP}{dt}}{P^2}

\dfrac{dV}{dt} = 8.31 \dfrac{18 \times 0.1-285\times 0.07}{18^2}

\dfrac{dV}{dt} = 8.31 \dfrac{-18.15}{324}

\dfrac{dV}{dt} = -0.466 m^3/s

7 0
3 years ago
The mass flow rate of air compressed in axial flow compressor is compressor a. lower than b. higher than c. same as d. unpredict
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8 0
3 years ago
When the same amount of heat is added to equal masses of water and
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