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swat32
3 years ago
13

Help me please,

Physics
1 answer:
garik1379 [7]3 years ago
7 0

Option B

Explanation:

no distance was given only the acceleration due to the fact that it went up (10m/s/s)

s0 it is

0 m/s and 10m/s/s (option B)

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say I have two rocks of different sizes but are taken from the same rock. Will their density be equalvilent
poizon [28]

yes their density are equal

8 0
3 years ago
If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second
Gre4nikov [31]

Answer: 500 s

Explanation:

Speed v is defined as a relation between the distance d and time t:

v=\frac{d}{t}

Where:

v=3(10)^{8}m/s is the speed of light in vacuum

d=1.5(10)^{11}m is the distance between the Earth and Sun

t is the time it takes to the light to travel the distance d

Isolating t:

t=\frac{d}{v}

t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}

Finally:

t=500 s

5 0
3 years ago
At this rate, how long would it take for two continents 3500 kilometers apart to collide?
meriva

<span>Given:

3,500 kilometers

Find:</span>

 

Years for two continents to collide = ?

 

<span>Solution:

We know that </span>typical motions of one plate relative to another are 1 centimeter per year.

So first, we convert 3,500 km to cm.<span>
</span><span>

</span>

The solution would be like this for this specific problem:

 

1 km = 100,000 cm

3,500 km x 100,000 = 350,000,000 cm

Since we know that 1 cm = 1 year, then that means 350,000,000 cm is equivalent to 350,000,000 years.

 

Therefore, it would take 350 million years for two continents that are 3500 kilometers apart to collide.

<span>
To add, </span>a phenomenon of the plate tectonics of Earth that occurs at convergent boundaries is called the continental collision.

5 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
3 years ago
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