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3 years ago

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The two basic units of weight in the metric system is the___?

1 answer:

Alika [10]3 years ago

3 0

Answer:

The two basic units of weight in the metric system is the?

The gram and kilogram is the Correct Answer...

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The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current

irina1246 [14]

Answer:

a average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

6 0

3 years ago

Read 2 more answers

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

A body travels at an initial speed of 2.5 m/s. Given a constant acceleration of 0.2 m/s 2 what is the speed of the body at time

garri49 [273]

Answer:

<u>We are given:</u>

u = 2.5 m/s

a = 0.2 m/s/s

t = 25 seconds

v = v m/s

<u>Solving for 'v':</u>

From the first equation of motion:

v = u + at

Replacing the values

v = 2.5 + (0.2)(25)

v = 2.5 + 5

v = 7.5 m/s

6 0

3 years ago

Marcus tells Sabrina that for the last few months, he has not been able to detect moderate heat or cold with his fingertips. Whi

hram777 [196]

Answer: The free ending nerves.

Explanation:

At our fingertips, we have a lot of sensory nerve endings, that give information about changes that occur at your skin.

Like touching something with a given texture, feeling pain, or noticing changes in temperature.

There are different types of nerve endings, particularly the ones responsible to detect pain, and temperature are the free nerve endings.

So Marcus may have the free nerve endings damaged.

4 0

3 years ago

A woman pushes an oak chest across an oak floor at a constant speed of 0.450 m/s. The chest has a mass of 40.0 kg, and the coeff

natima [27]

Answer:

A. 243 N

Explanation:

Friction is the force that opposes the relative motion between systems that are in contact.

This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.

First find the normal force which is the force that would point directly upwards to support weight of the block.

Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.

Given m=40 kg and g=9.80 m/s²

N force=40×9.80 =392N

Then find the force of friction which is given by the formula;

<em>F=μN where μ is friction coefficient for the oak chest and N is the normal force on the chest</em>

Given <em>μ</em>=0.620 and N force = 392 N then it will be;

F=0.620× 392 =243.04 N

Answer : 243 N

6 0

3 years ago