All categories

3 years ago

What factors affect potential energy

Physics

2 answers:

jok3333 [9.3K]3 years ago

6 0

Mass ,gravity and height

Ugo [173]3 years ago

3 0

Gravitational potential energy is determined by three factors : mass, gravity and height . All three factors are directly proportional to energy

You might be interested in

Need help asap

Vanyuwa [196]

I’ve done this before the answer is B

8 0

1 year ago

The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta

ivanzaharov [21]

Answer:

(a) 11.3 L

(b) 10 M

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

Luminosity of Star = 11.3 L

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

Mass of Star = 10 M

3 0

3 years ago

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

4 0

3 years ago

An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago

Please answer D in the image with an explanation

puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

$T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}$

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0

2 years ago