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Marina CMI [18]
2 years ago
15

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the componen

t decreases to the half of it former value. What change will it occur in the current through it?
Physics
2 answers:
slamgirl [31]2 years ago
8 0

Answer:

The current through it will also decrease to half of its former value because according to Ohm's law the current flowing through a resistor is directly proportional to the potential difference applied across its ends provided that the temperature and some other necessary conditions remain constant.

This is mathematically represented as follows;

V=IR.........(1)

The current is thus given as

I=\frac{V}{R}..............(2)

if R is constant and V is reduced to half, then we have the following;

I=\frac{V/2}{R}

Simplifying further we obtain

I=\frac{V}{2R}...........(3)

Equation (3) shows that the current I is also reduced to half.

Illusion [34]2 years ago
5 0

Answer:

The current will be decreased to half of it former value(I/2)

Explanation:

According to Ohm’s law  which states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R).

V = IR

⇒ I=V/R ---------- (1)

Now Potential difference is decreased to half

∴ New potential difference Vʹ=V/2

Resistance remains constant

So the new current Iʹ = Vʹ/R

                          I¹  = (V/2)/R

                            I¹= (1/2) (V/R)

                           I¹ = (1/2) I = I/2

The current will be halved.

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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
2 years ago
You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work
serious [3.7K]

Answer:

Check the explanation

Explanation:

1) Pressure acting on the plug = Patm + P

Pressure = Patm + rho*g*h (Here h = D2)

Pressure = 101325 + 1000*9.8*7

Pressure = 169925 Pa

so, Force = PA

Force = 169925*pi*0.0152

Force = 120.1 N

7 0
3 years ago
As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

3 0
3 years ago
Which of these reactions is responsible for energy radiating from stars, including the Sun?
snow_tiger [21]

Answer: nuclear fusion

Explanation: give me a brainliest

3 0
2 years ago
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