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2 years ago

What is the definition of a reference point

Physics

1 answer:

riadik2000 [5.3K]2 years ago

8 0

A reference point would be something used to understand or judge or describe a certain topic in discussion.

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A "590-W" electric heater is designed to operate from 120-V lines.

lukranit [14]

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

I = 4.92 A

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

R = 24.4 Ω

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

P = 495.9 W

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

7 0

2 years ago

What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?

Igoryamba

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

5 0

2 years ago

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Which of these habitats has a highly salinity

Llana [10]

The answer for APEX Learning is Ocean

7 0

2 years ago

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The universal force that is effective over the longest distance is called?

gulaghasi [49]

The answer is D. gravity

3 0

2 years ago

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

6 0

2 years ago