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Aloiza [94]
3 years ago
7

Which one has greater potential energy? An 8 kg rock sitting on a 2.2 m cliff or a 6 kg rock sitting on a 3.2 m cliff.

Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0

Answer: The 6 kg rock sitting on a 3.2 m cliff.

Explanation:

The potential energy of an object of mass M that is at a height H above the ground us:

U = M*H*g

where g is the gravitational acceleration:

g = 9.8m/s^2

Then:

"An 8 kg rock sitting on a 2.2 m cliff"

M = 8kg

H = 2.2m

U = 8kg*2.2m*9.8 m/s^2 = 172.48 J

"a 6 kg rock sitting on a 3.2 m cliff"

M = 6kg

H = 3.2m

U = 6kg*3.2m*9.8m/s^2 = 188.16 J

You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.

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7 0
3 years ago
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

7 0
3 years ago
Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k
Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

\frac{GM_{ns}}{R^2}= \omega^2 R

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0
3 years ago
Water that flows from behind a large dam can cause machines to produce electricity. What change takes place?
lyudmila [28]
The mechanical energy in the falling water is used to spin the generator, and gets transformed into electrical energy.  That's the first choice on the list.
8 0
3 years ago
The car starts from a room with a constant acceleration of 5 ms-2. What path will it pass in 6 seconds and what speed will it re
Kobotan [32]

Answer: The speed will be  30 m/s .

Explanation:

Given: Initial velocity of the car: u = 0 m/s

Constant Acceleration: a = 5 m/s²

Time: t= 6 seconds

To find: Final velocity(v)

Formula:  v = u+at

Substitute values in the formula, we get

v=  0+(5)(6) m/s

⇒ v= 30 m/s

i.e. Final velocity = 30 m/s

Hence, the speed will be 30 m/s .

4 0
2 years ago
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