Plzzzz help meeeeeeee

Answer:

I think it would be c, crashing into another car

The radius of the satellite's motion is 2.49 × 10⁷ m

Since the centripetal force, F = mv²/r equals the gravitational force, F' = GMm/r² on the satellite,

F = F'

mv²/r = GMm/r²

v² = GM/r

Making r subject of the formula, we have

r = GM/v² where r = radius of satellite's motion, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.972 × 10²⁴ kg and v = speed of satellite = 4,000 m/s.

Substituting the values of the variables into the equation, we have

r = GM/v²

r = 6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg/(4,000 m/s)²

r = 39.83324 × 10¹³ Nm²/kg ÷ 16 × 10⁶ m²/s²

r = 2.4895775 × 10⁷ m

r ≅ 2.49 × 10⁷ m

So, the radius of the satellite's motion is 2.49 × 10⁷ m.

Learn more about orbital radius of satellite motion here:

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Answer:

25.0 million meters

Explanation:

Answer:

Object distance should be 48

Explanation:

in pic

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Answer:

Cellular respiration is the process in which cells break down glucose, release the stored energy, and use the energy to make ATP.

Explanation:

Answer: One rocket has more fuel than the other

Two rockets are being tested. Both rockets have the same mass. One rocket accelerates at a higher rate than another rocket. One rocket has more fuel than the other

A rocket is a special type of jet-propulsion device which carry either solid or liquid propellants and it provide both the fuel and oxidizer needed for combustion.

The rocket is commonly applied to various vehicles, including firework skyrockets, guided missiles, and launch vehicles applied in spaceflight..

The rocket is different from the turbojet and other “air-breathing” engines exhaust jet consists of the gaseous combustion products of “propellants” carried on board.

The  physical principle involved in rocket propulsion was  derived by Sir Isaac Newton. According to third law of motion, the rocket experiences an increase in momentum proportional to the momentum carried away in the exhaust,

For more details regarding rocket , visit

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Hi there!

Since we are dealing with such large distances, we can approximate.

Recall that the equation for speed is:

displacement / time = velocity

OR:

d/t = v

Begin by converting one hour to seconds:

1 hr  = 3600 sec

Now, we can solve for velocity:

(-2.0 × 10⁷) / 3600 = -5555.6 m/s ⇒ D: -5.55 × 10³ m/s

Answer:

Friction or tension

Explanation:

Friction stops an object from moving in the presence of force

Hi there!

A)

The angle that will produce the same range is the compliment of 35°.

Thus, kicking the ball at 55° will result in the same range.

We can prove this by using the derived range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

An angle of 35° yields:

[tex]R = \frac{v^2sin(2*35)}{g} = .939R[/tex]

An angle of 55° yields:

[tex]R = \frac{v^2sin(2*55)}{g} = .939R[/tex]

Both are the same, thus indicating that 55° produces the same range.

B)

The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.

sin(55) > sin(35)

C)

The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.

We can prove this using:

vf = vi + at

0 = vy - 9.8t

vy/9.8 = t

Greater vy (vertical component of velocity) ⇒ greater time taken.

D)

The angle that would result in the furthest range is 45°.

We can prove this using calculus. Recall the above range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

We can take the derivative and use the first-derivative test to find its critical point:

[tex]\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0[/tex]

Evaluate:

[tex]v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}[/tex]

A. The angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

A. The range of a projectile is the horizontal distance it travels before it hits the ground. The range of a projectile is determined by the initial velocity, the angle of projection, and the acceleration due to gravity.

For a given initial velocity, the range of a projectile is maximized when the angle of projection is 45°. However, if the ground is not level, the range of a projectile can be maximized at other angles.

In the case of a soccer ball kicked at an angle of 35°, the range will be maximized at an angle of 53°. This is because the range of a projectile is maximized when the vertical component of the initial velocity is equal to the horizontal component of the initial velocity.

The angle of 53° is the angle that produces a vertical component of the initial velocity that is equal to the horizontal component of the initial velocity when the ball is kicked at an angle of 35°.

Therefore, the angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. The angle that produces the furthest range depends on the initial velocity and the acceleration due to gravity. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

To know more about angle here

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Hi there!

We can begin by calculating the impulse exerted on the soccer ball:

I = Δp = mΔv

Plug in the given values:

I = 0.45 × 20 = 9 Ns

I = Ft, so:

I/t = F

9/0.1 = 90 N

Answer:

Oh wow, that's a long answer. I don't know what voltage is on the circuit.

Answer:

101.937 kg

Explanation:

The force needed to get the box moving must just cancel the static friction force:

F = µsmg  =  0.35•m•9.81 = 350 --->   m = 350 / (0.35•9.81) = 101.937 kg

Again, with units shown, and using 1 N = 1 kg•m/s2:

0.35•m•9.81(m/s2) = 350 N

solving for m:

m = 350 N / (0.35•9.81 m/s2)  =  (350 kg•m/s2) / (3.434 m/s2 ) = 101.937 kg

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