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3 years ago

In a position-time graph what does it mean when you return to 0? ASAP HELP PLEASE!

Physics

1 answer:

Novosadov [1.4K]3 years ago

6 0

It means that you have returned to the designated starting position.

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Granite is uplifted by the movement of tectonic plates. It will most likely become _____. igneous rock sedimentary rock metamorp

pashok25 [27]

Answer:

The answer would be Igneous rock

Explanation:

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2 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°

dezoksy [38]

fdi9bn 08989089089008i uuuuuuri9ij

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3 years ago

Does kinetic energy stay the same at all heights? pls help I have 12 minutes to finish the project and the teacher won't help me

Alex Ar [27]

Answer:

Kinetic energy does not stay the same at all heights

Explanation:

Well as the height and wind increase so does the kinetic energy it's like when you fall as you are about to hit the floor you speed increases

HOPE THIS HELPS YA :)

7 0

2 years ago

If a 20 kg green fish swimming at 2 m/s swallows a 1 kg orange fish at rest, in what direction, and how fast

krok68 [10]

Answer: 1.9 m/s

Explanation:

The question should be:

If a 20 kg green fish swimming at 2 m/s swallows a 1 kg orange fish at rest, in what direction, and how fast will the green fish swim after eating the orange fish?

Ok, here we have conservation of momentum.

At the beginning, the total momentum is equal to the sum between the momentum of the green fish and the momentum of the orange fish.

Where the momentum is written as:

P = m*v

m = mass

v = velocity.

The momentum of the green fish is:

Pg = 20kg*2m/s = 40 kg*m/s.

The momentum of the orange fish is:

Po = 1kg*0m/s = 0

The total initial momentum is:

Pi = Pg + Po = 40 kg*m/s.

After the green fish eats the orange fish, we do not have an orange fish anymore, and the mass of the green fish will be equal to it's initial mass, plus the mass of the fish that it ate, this will be:

M = 20kg + 1kg = 21kg.

Then the momentum will be:

Pf = 21kg*V

Where V is the final velocity.

For conservation of momentum, the initial momentum is equal to the final momentum, then:

Pi = Pf

40 kg*m/s = 21kg*V

(40/21) m/s = 1.9 m/s = V

The fish's final velocity is 1.9 m/s

5 0

2 years ago