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Rudiy27
4 years ago
7

A space probe is fired as a projectile from the earth's surface with an initial speed of 2.44 104m/s. what will its speed be whe

n it is very far from the earth? ignore atmospheric friction and the rotation of the earth.
Physics
1 answer:
Lostsunrise [7]4 years ago
8 0
Projectile motion is characterized by an arc-shaped direction of motion. It is acted upon by two vector forces: horizontal component and vertical component. The horizontal component is in constant velocity motion, while the vertical component is in constant acceleration motion. These two motions are independent of each other.

Now, the total velocity of the space probe at the end of the projectile motion is determined through this equation:

V = √(Vx² + Vy²)

where Vx is the velocity in the horizontal direction and Vy is the velocity in the vertical direction.

Let's find Vx first. Assuming that the space probe was launched at an angle horizontal the Earth's surface, the launching angle is 0°. Thus, the initial velocity is 2.44×10⁴ m/s. 

For Vy, the free falling motion is 
Vy = √(2gh), where g is 9.81 m/s² and h is the distance traveled vertically by the space probe which represents the radius of the Earth equal to 6.37×10⁶ meters. Therefore,

V = √{(2.44×10⁴)² + [√(2×9.81×6.37×10⁶ )]²}
V = 26,839.14 m/s
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3 years ago
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If a CFOC was launched and travels 65 meters and is in the air for 3 seconds, what is the launch velocity and angle?
labwork [276]

Answer:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

Explanation:

From the question given above, the following data were obtained:

Range (R) = 65 m

Time of flight (T) = 3 s

Acceleration due to gravity (g) = 10 m/s²

Lauch velocity (u) =?

Lauch Angle (θ) =?

R = u²Sin2θ /g

65 = u² × Sin2θ /10

Recall:

Sin2θ = 2SinθCosθ

65 = u² × 2SinθCosθ / 10

65 = u² × SinθCosθ / 5

Cross multiply

65 × 5 = u² × SinθCosθ

325 = u² × SinθCosθ .....(1)

T = 2uSinθ / g

3 = 2uSinθ / 10

3 = uSinθ / 5

Cross multiply

3 × 5 = uSinθ

15 = u × Sinθ

Divide both side by Sinθ

u = 15 / Sinθ....... (2)

Substitute the value of u in equation (2) into equation (1)

325 = u² × SinθCosθ

u = 15 / Sinθ

325 = (15 / Sinθ)² × SinθCosθ

325 = 225 / Sin²θ × SinθCosθ

325 = 225 × SinθCosθ / Sin²θ

325 = 225 × Cosθ / Sinθ

Cross multiply

325 × Sineθ = 225 × Cosθ

Divide both side by Cosθ

325 × Sineθ / Cosθ = 225

Divide both side by 325

Sineθ / Cosθ = 225 / 325

Sineθ / Cosθ = 0.6923

Recall:

Sineθ / Cosθ = Tanθ

Tanθ = 0.6923

Take the inverse of Tan

θ = Tan¯¹ 0.6923

θ = 35°

Substitute the value of θ into equation (2) to obtain the value of u.

u = 15 / Sinθ

θ = 35°

u = 15 / Sin 35

u = 15 / 0.5736

u = 26.15 m/s

Summary:

Lauch velocity (u) = 26.15 m/s

Lauch Angle (θ) = 35°

8 0
3 years ago
Which statements describe a situation with a displacement of zero? Check all that apply.traveling south for 30 miles, then turni
cluponka [151]

Answer:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

Explanation:

Displacement simply means the.change in position of an object. In a situation whereby the initial and final position are thesame, the displacement will be zero.

The statements that describe a situation with a displacement of zero include:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

5 0
3 years ago
A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find
Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

8 0
3 years ago
HURRY!!!!!!!!!!!
iren2701 [21]

The density of a substance is the ratio of its mass to its volume. So whatever volume you have (call it V), the density \rho is computed by

\rho=\dfrac{200\,\mathrm g}V

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