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3 years ago

The earth rotates once per day about an axis passing through the north and south poles. True or False

Physics

1 answer:

alexandr402 [8]3 years ago

5 0

Answer: False.

Explanation: The Earth rotates once per day in direction from East to West. The rotation of the Earth daily is responsible for day and night experienced, the areas of the Earth facing the Sun experiences day time while the areas of the Earth away from the Sun in the Earth's shadow experiences night time.

While the revolution of the Earth around the gives rise to a year, as it takes 365 days for the Earth to go round the sun.

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In a fission experiment observed by Hahn and Strassman, uranium-235 was bombarded by neutrons. The products of this ission react

wlad13 [49]

Answer:

helium-4 (90%) or tritium (7%).

Explanation:

hope it helped u buddy

7 0

3 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

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

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

3 years ago

He generation of a magnetic field by an electric current is

USPshnik [31]

Answer: electromagnetism.

Explanation:The use of coils of wires produces a relationship between electricity and magnetism that gives us another magetism called electromagnetism.

6 0

3 years ago

What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

azamat

Explanation:

Q1) What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

Ans1) speed, v=st=2πrT=2×227×7×10-260×60=119×10-4=1.22×10-4m/s

<h2><em><u>Hope it helps</u></em></h2>

5 0

3 years ago