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Assoli18 [71]
3 years ago
8

A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of

gravity on this satellite will double but the satellite's orbit will not be affected. A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. True False
Physics
2 answers:
adelina 88 [10]3 years ago
7 0

Answer:

it is True

Explanation:

I hope this helps

Fofino [41]3 years ago
5 0

Answer:

Explanation:

True, provided that the extra mass is also already in the same orbit itself.

If the extra mass is NOT in the same orbit, it's called a collision and the orbits of both will be altered due to the changes in momentum and kinetic energy.

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James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s
liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{James}=\sqrt{\frac{2h}{g}}

Vertical motion of John : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{John}=\sqrt{\frac{2h}{g}}

So both times are same.

Both of them reach the lake at the same time.

3 0
3 years ago
A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car
jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

8 0
2 years ago
If two opposite charges get farther apart, does their attraction for each other increase, decrease, or remain the same?​
wlad13 [49]

The force between them <em>decreases</em><em>,</em> as the square of the distance.

For example ...

-- If you move them apart to double the original distance, the force becomes (1/2²) = 1/4 of the original force.

-- If you move them apart to 3 times the original distance, the force becomes (1/3²) = 1/9 of the original force.

-- If you move them apart to 5 times the original distance, the force becomes  (1/5²) = 1/25 of the original force.

(Gravity works exactly the same way.)

7 0
3 years ago
A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the t
Virty [35]

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

          Em₀ = U = m g h₁

Final point. Lower (slide bottom)

            Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

             mgh₁ = ½ m v² + mgh₂

             v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

           x = v₀ₓ t

          y = v_{oy} t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

         t = √ 2h₂ / g

We substitute in the other equation

         d = √ (2g (h₁-h₂))  √ 2h₂ / g

         d = √ (4 h₂ (h₁-h₂))

         H = h₁ + h₂

         h₁ = H -h₂

         d = √  ( 4 h₂ ( H - 2h₂))

3 0
3 years ago
Read 2 more answers
Help asap plz !!! Use the circuit diagram to decide if the lightbulb will
nata0808 [166]

Answer:

I wish I knew the answer to that question

4 0
3 years ago
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