All categories

3 years ago

A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

Physics

2 answers:

love history [14]3 years ago

6 0

Answer:

$R\approx12.43 \,\, \Omega$

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

$V=I\,\,R\\R = \frac{V}{I} \\R=\frac{87}{7} \\R\approx12.43 \,\, \Omega$

kramer3 years ago

3 0

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

I would appreciate it!

You might be interested in

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o

dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

$F\propto\frac{d}{dt} (p)$

$\Rightarrow F=\frac{d}{dt} (m.v)$

since mass is constant

$F=m\frac{d}{dt}v$

when $dt=10t$

then,

$F'=m.\frac{v}{10\times t}$

$F'=\frac{1}{10} \times \frac{m.v}{t}$

$F'=\frac{F}{10}$ the body will experience the tenth part of the maximum force.

where:

$\frac{d}{dt} =$ represents the rate of change in dependent quantity with respect to time

$p=$ momentum

$m=$ mass of the person jumping

$v=$ velocity of the body while hitting the ground.

7 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

What is universal laws of gravitation

valentinak56 [21]

Answer:

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Explanation:

7 0

3 years ago

Is the earth's gravitational force on the sun larger than, smaller than, or equal to the sun's gravitational force on the earth?

Leona [35]

Answer:

The earth's gravitational force on the sun is equal to the sun's gravitational force on the earth

Explanation:

Newton's third law (law of action-reaction) states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In other words, when two objects exert a force on each other, then the magnitude of the two forces is the same (while the directions are opposite).

In this problem, we can call the Sun as "object A" and the Earth as "object B". According to Newton's third law, therefore, we can say that the gravitational force that the Earth exerts on the Sun is equal (in magnitude, and opposite in direction) to the gravitational force that the Sun exerts on the Earth.

6 0

3 years ago