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3 years ago

Drew wants to convert square mils to circular mils in an area calculation. How does he do this?

Physics

1 answer:

bija089 [108]3 years ago

6 0

Answer:

i dont know

Explanation:

blah blah blah your welcome

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PLEASE HELP!! 50 PTS!!

eduard

Answer

The direction of the sun’s pull relative to the planet’s motion changes.

Explanation:

4 0

3 years ago

Read 2 more answers

Which car is accelerating?

Oliga [24]

Answer:

The car that is accelerating is B a car that rounds a curve at a constant speed

Explanation:

Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.

8 0

2 years ago

Read 2 more answers

a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(1

tekilochka [14]

By Newton's second law, the net force on the object is

∑ F = m a

∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N

Let f be the unknown force. Then

∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f

=> f = (-2.0 i - 12.0 j ) N

8 0

3 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 force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball

shepuryov [24]

Explanation:

Hello there!!!

You just need to use simple formula for force and momentum, 

F= m.a

and momentum (p)= m.v

where m= mass

v= velocity.

a= acceleration .

And the solutions are in pictures. 

Hope it helps..

5 0

3 years ago