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2 years ago

9

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

tric field 4 cm from the plate is:

1 answer:

AysviL [449]2 years ago

7 0

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

2 years ago

How much power does a machine have that does 15204 J of work in 40 seconds?

pentagon [3]

Explanation:

since P=W/t

P=15204/40

P=380.1 Watt

5 0

1 year ago

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

$F=ma$

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0

2 years ago

Read 2 more answers

On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo

timofeeve [1]

Answer:

a = 0.5 m/s²

Explanation:

Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:

ωf = ω₀ + α*t

⇒ ωf = α*t ⇒ α = $\frac{wf}{t}$ = $\frac{1.4 rad/s}{21 s} = 0.067 rad/s2$

The angular velocity, and the linear speed, are related by the following expression:

v = ω*r

Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²

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2 years ago

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.

shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

$F = \mu W$

we know that F = ma

$ma = \mu (2/3 mg)$

$a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2$

the new power is $p = 70\% P_[max} = 0.7 P_{max}$

$P =f_{max} v$

$0.7P_{max} = ma_{max} v$

solving for speed v

$v =0.7 \times \frac{P_{max}}{ma_{max}}$

$v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}$

v = 11.5689 m/s

7 0

2 years ago