All categories

3 years ago

A wire with a resistance of 20 Ω is connected to a 12-V battery. What is the current flowing through the wire?

Physics

2 answers:

erma4kov [3.2K]3 years ago

8 0

Current is 0.6 Amps i think

Semenov [28]3 years ago

3 0

Answer : Current flowing through the wire is 0.6 A

Explanation :

Given that,

Resistance of the wire, $R=20\ \Omega$

Voltage, $V=12\ V$

Using Ohm's law :

$V=IR$

$I=\dfrac{V}{R}$

$I=\dfrac{12\ V}{20\ \Omega}$

$I=0.6\ A$

Hence, the current flowing through the wire is 0.6 A

You might be interested in

A man on a bicycle of total mass of 10kg has a Velocity of 2 ms.' He Paddles faster for 5 seconds and the velocity Increase to i

sesenic [268]

Answer:

the value of force, F=4.0N

Explanation:

Firstly, recall velocity-time equation

v=u+at
(4)=(2)+a(5)
a=0.4m/s²

Secondly, recall the Newton's 2nd Law

F=ma
F=(10)(0.4)
F=4.0N

7 0

1 year ago

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago

Please need help on this

neonofarm [45]

Answer:

Explanation:

Geothermal energy is heat driven within the sub-surface of the earth. Water and/or steam carry the geothermal energy to the earth surface.

6 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

(6) Module 4 Study Guide

velikii [3]

We have the meats Arby’s we beat them kids

3 0

3 years ago