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3 years ago

What is an electron, where is it found?

Physics

1 answer:

Anna11 [10]3 years ago

5 0

A electron is a particle with negative charge and it surrounds the nucleus of an atom

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The moon Phobos orbits Mars

shepuryov [24]

$27.9816 \times 10^{3} s$ is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

$\frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}$

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant $6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$

Here, given data

$M=6.23 \times 10^{23} \mathrm{kg}$

$R=9.38 \times 10^{6} \mathrm{m}$

Substitute the given values, we get T as

$\frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}$

$T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}$

$T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}$

Taking square root, we get

$T=27.9816 \times 10^{3} s$

4 0

3 years ago

What is the force when an object exerts 15 pascal pressure on a square surface of side 5 meters

olga55 [171]

Answer:375N

Explanation:

pressure =15 pascal

Length of side =5 m

Area=length x length

Area=5 x 5

Area=25 m^2

Force=pressure x area

Force=15 x 25

Force=375N

3 0

3 years ago

It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

trapecia [35]

Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

Rate of Re crystallization at this temperature

$t^{-1}=8.76\times 10^{-3} min^{-1}$

3 0

3 years ago

Select all the correct answers.

velikii [3]

Soda and metal, so b and c :)

8 0

3 years ago

Read 2 more answers

While playing catch with my dog I bounced the ball off the ground at a 30-degree angle. It had a range of 6 meters.

baherus [9]

The initial velocity of the ball is 8.2 m/s

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A uniform motion along the horizontal direction, at constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration ( $g=9.8 m/s^2$ , acceleration of gravity)

The range of a projectile can be derived by the equation of motions along the two directions, and it is found to be:

$d=\frac{v^2 sin 2\theta}{g}$

where

v is the initial velocity of the projectile

$\theta$ is the angle of projection

g is the acceleration of gravity

For the ball in this problem, we have

$\theta=30^{\circ}$

d = 6 m is the range

Solving for v, we find the initial velocity:

$v=\sqrt{\frac{gd}{sin 2\theta}}=\sqrt{\frac{(9.8)(6)}{sin (2\cdot 30^{\circ})}}=8.2 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago