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1 year ago

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Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measur

e of the electric force on any charged object. It is a measure of the ratio of the charge on an object to its mass. It is a measure of the electric force per unit mass on a test charge. It is a measure of the electric force per unit charge on a test charge. It is a measure of the total charge on the object.

1 answer:

IceJOKER [234]1 year ago

5 0

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

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A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

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2 years ago

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

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Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

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2 years ago

During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)

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T= 24.5 feet per second. That is the velocity it reaches at the end of its fall

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2 years ago

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Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

konstantin123 [22]

Answer: His average speed in mph over the last 400 m is 7.7 m/s.

Explanation:

Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

We know that , speed = $\dfrac{distance}{time}$

Here , distance = 400m and time = 51.9 s

Then, speed = $\dfrac{400}{51.9}\approx7.7\ m/s$

Hence, his average speed in mph over the last 400 m is 7.7 m/s.

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2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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2 years ago