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earnstyle [38]
3 years ago
10

Describe the relationship between motion and a reference point.

Physics
2 answers:
Stells [14]3 years ago
8 0

Answer:

A <u>object </u>is in motion when its distance from another object it changing.

a <u>reference </u>point is a place or object used for comparison to determine if something is in motion.  You can assume that the refrence point is stuck or not moving.

steposvetlana [31]3 years ago
4 0

Answer:

An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.

Explanation:

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andriy [413]
Salt is a substance because a substance is a material that has constant composition regardless of its source. Salt is the only option that matches this description
5 0
3 years ago
A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds
Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

   = ∫(12 + 5×t)×dt

   = 12×t + 2.5×t^2

then:

Ф = ∫W×dt

   = ∫(12×t + 2.5×t^2)dt

   = 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

   = 239.73 rad    

5 0
3 years ago
Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120
igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter. 

As you have the circumference of B, you divide it by Pi to get the Diameter. 

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter. 

As' radius and Diameter will be 3x greater than B. 

38.2 x 3 = 114.6 

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 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
3 years ago
It is known that heat is added to a gas in sealed container. The container is fitted with a moveable piston.
jasenka [17]

Answer:

d. Not enough information is given to answer this question.

Explanation:

From first law of thermodynamics

Q= W + ΔU

Q=Heat  ,W= Work , ΔU=Change in internal energy

If work done by the gas :

It means that W and Q both are positive

Q- W = ΔU

Ii Q > W ,then temperature of the gas will increase.

If  Q< W  ,Then temperature of the gas will decreases.

If work done on the gas:

Q positive but W will be negative

Q- W = ΔU

Q= W or Q>W  or Q< W ,then temperature of the gas will increase.

There are three cases because they did not give any information about the work.That is why option d is correct.

3 0
3 years ago
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