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3241004551 [841]
3 years ago
7

When an object is in orbit, it is falling at the same rate at which the Earth is curving. Please select the best answer from the

choices provided
a. True
b. False
Physics
1 answer:
olga2289 [7]3 years ago
7 0

The statement “When an object is in orbit, it is falling at the same rate at which the Earth is curving” is true. The speed of a satellite orbiting the earth depends only on the mass of the earth and the mass of the satellite.

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A train travelling at 20 m/s has 2 000 000 J of kinetic energy. What is the mass of<br> the train?
Svetlanka [38]

Answer:

<h2>10,000 kg</h2>

Explanation:

The mass of the train can be found by using the formula

m =  \frac{2k}{ {v}^{2} }  \\

k is the kinetic energy

v is the velocity

From the question we have

m \frac{2(2000000)}{ {20}^{2} }  =  \frac{4000000}{400}  =  \frac{40000}{4}  \\

We have the final answer as

<h3>10,000 kg</h3>

Hope this helps you

3 0
3 years ago
Read 2 more answers
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
3 years ago
The stars in the sky are organized into groups of stars called constellations which appear near each other in the sky but are no
bezimeni [28]

Answer:

The International Astronomical Union (IAU)  has accepted 88 constellations in the sky.

Explanation:

Constellations has been used since the beginnings of civilizations and each one of them named them as they considered appropiate. It means Greeks' constellations were different than the ones described by Chinese, so it was necessary to gather all these constellations and make a great record with all of them, but there was a problem: Some constellations from different civilizations overlaped because they shared the same stars. There was necessary to put some order on this and that is when in 1922 the International Astronomical Union (IAU) defned a set of 88 moderm constellations  that would become the international standard to look at the night sky. Each one of them is unique and does not share stars with the other constellations.  

6 0
3 years ago
A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) = B0 cos(ωt) ˆz.
Alexxandr [17]

Answer:

Explanation:

Given that,

B(t) = B0 cos(ωt) • k

Radius r = a

Inner radius r' = a/2 and resistance R.

Current in the loop as a function of time I(t) =?

Magnetic flux is given as

Φ = BA

And the Area is given as

A = πr², where r = a/2

A = πa²/4

Then,

Φ = ¼ Bπa²

Φ(t) = ¼πa²Bo•Cos(ωt)

Then, the EMF is given as

ε(t) = -dΦ/dt

ε(t) = -¼πa²Bo • -ωSin(ωt)

ε(t) = ¼ωπa²Bo•Sin(ωt)

From ohms law,

ε = iR

Then, i = ε/R

I(t) = ¼ωπa²Bo•Sin(ωt) /R

This is the current induced in the loop.

Check attachment for better understanding

7 0
3 years ago
Can someone please Help me with this? It’s Due today
kirza4 [7]

helium group no val elect

mg is reactive when activated. when burned, very intense

pot\asssium 1 valence elect ... KCl eg

theone with H and sodium in it

http://perendis.webs .com

6 0
3 years ago
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