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3 years ago

The unit of electric potential or electromotive force is the

Physics

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

Volt

Explanation:

Electric Potential is the Voltage and it's measured in volt

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In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago

A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal

sammy [17]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

8 0

3 years ago

If the change in kinetic energy of a tennis ball hit by the racket

Marrrta [24]

Answer: .36 m

Explanation:

5 0

3 years ago

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a

Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0

3 years ago

During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a

lana66690 [7]

According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.

8 0

3 years ago