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PIT_PIT [208]
3 years ago
15

Four resistors are wired in series with a 120 V source, with values of 8.0 ohms, 12 ohms, 14 ohms, and 16 ohms. What is the tota

l current in the circuit?
A. 2.4 A
B. 3.0 A
C. 2.0 A
D. 7.5 A

PLEASE HELP. NO LINKS.
Physics
1 answer:
vovangra [49]3 years ago
4 0

Answer:

Resistors in series

Equivalent Resistance = 8+12+14+16=50ohms

V=IR

I = V/R

= 120/50

=2.4A

OPTION A

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DanielleElmas [232]

The options are missing and they are;

Select one:

A. By seeing it flying above the water

B. By listening for it in the air

C. By listening for it underwater

D. None of the above: it won’t be able to detect it in time.

Answer:

OPTION C is the correct answer:By listening for it underwater

Explanation:

Due to the fact that the ships and other boats have underwater microphones because on most cargo ships the engine is to loud so they use microphones underwater to detect torpedos. Thus, the vessel can detect the torpedo by listening under water

8 0
3 years ago
Can someone help me with this please
rjkz [21]

Answer:

1. 2HCl + MgO —-> MgCl2 + H2O

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4 0
3 years ago
When a 20 V battery is hooked up to a 100 Ω resistor in a circuit, how much current flows through the circuit?
horsena [70]
From the equestion:
I=V÷R
I=20÷100
I=0.2A
6 0
3 years ago
The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s
JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

a)

We will use second equation of motion to find the height:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\

<u>h = 593.50 m</u>

b)

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height covered during last second = h₁₁ =  ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\

<u>h₁₁ = 103 m</u>

c)

Now, we use first equation of motion for complete motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)

<u>vf = 107.91 m/s</u>

8 0
4 years ago
a 5kg block on a rough horizontal surface is attached to a light spring (force constant=1.6kN/m). the block passes through its e
natima [27]

Answer:

2.12 J

Explanation:

Initial kinetic energy = final elastic energy + work by friction

KE = EE + W

KE = ½ kx² + W

5 J = ½ (1600 N/m) (0.06 m)² + W

W = 2.12 J

5 0
3 years ago
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