Ask question

Ask question

All categories

2 years ago

5

a lunar probe is moving 1670 m/s at a 73.0 angle. it needs to land on the moon 3.71 x 10^8 m away in a 46.0 direction in 8.64 x

10^4 s. what is the magnitude of the acceleration that the engine must produce?

1 answer:

Oduvanchick [21]2 years ago

8 0

The magnitude of the acceleration that the engine must produce is 0.065 m/s².

<h3>What is acceleration?</h3>

Acceleration is given by the ratio of Resultant or total force acting on any object and the its mass.

It can also be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given is a lunar probe is moving 1670 m/s at a 73° angle. it needs to land on the moon 3.71 x 10⁸ m away in a 46° direction in 8.64 x 10⁴ s.

The initial velocity will be

1670 x cos (73°-46°) = 1487.981 m/s

Using the second equation of motion, we have

s = ut + 1/2 at²

The final velocity will be zero after landing on Moon.

Substituting the values, we have

3.71 x 10⁸= 1487.981 x 8.64 x 10⁴ + 1/2 a ( 8.64 x 10⁴)²

Solving the equation, we get

acceleration, a = 0.065 m/s²

Thus, the magnitude of the acceleration that the engine must produce is 0.065 m/s²

Learn more about acceleration.

brainly.com/question/12550364

#SPJ1

You might be interested in

A car is driving along a road at a speed of 50 km/hr. How far will the car travel in 5 hours?

BARSIC [14]

Answer:

I think it's 250

Explanation:

If the car is traveling 50 km/hr that means every hour, the car drives 50 km. So if you want to know how far it will go in 5 hours you do 50x5.

4 0

3 years ago

What does an unconformity in a sequence of rock layers reveal about geologic history?

andreev551 [17]

An unconformity, or abnormality in the layers of rocks, signifies that the process of layering was somehow interrupted. The only feasible choice for this would be (B), where erosion would have interrupted the layering process.

7 0

3 years ago

Read 2 more answers

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

4 years ago

A construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity if he travels a distance o

Anastasy [175]

Answer:

W = 0

Explanation:

We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.

The work done by an object is given by :

$W=Fd$

F = ma

So,

$W=mad$

m is mass

a is acceleration

d is displacement

The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.

8 0

3 years ago

Describe and compare the rings of Saturn and Uranus, including their possible origins.

Marta_Voda [28]

Explanation:

Okay, well, Saturn's rings form a wide and complex system, consisting mostly of particles and pieces of ice, and are highly visible. They may have formed from one or more moons that broke up due to a collision, or are left over from early debris that never coalesced into a moon... And, The rings of Uranus are thin and hard to see, consisting mostly of chunks of carbon and hydrocarbons with very little reflectivity. They may also have formed from the breakup of a small moon due to a collision. They may be kept thin by the presence of shepherd moons.

Hope I helped !

:)

6 0

3 years ago