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3 years ago

If we find v=Al, where l is a length and v is a speed, what are the SI units for A?

Physics

2 answers:

12345 [234]3 years ago

8 0

Answer:

A) S^-1

Explanation:

Hope I helped

If there is any thing feel free to ask

You are welcome in advance

swat323 years ago

3 0

Answer:

I think the SI unit of A is m^2/s

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The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately 0.199 rad/s

Second question:

The telescope should accelerates away by approximately F = 0.0005·m

Third question:

The pulling force between the Earth and the satellite

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ 0.199 rad/s

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

The force of attraction between the Earth and the telescope which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

3 0

3 years ago

The exact speed of an object in a specific instant is?

GarryVolchara [31]

Im not so sure but it should be the
instantaneous speed

3 0

3 years ago

Read 2 more answers

Can someone please help me no one ever responds to my question and I need the correct answer. I believe the answer is C but i'm

GarryVolchara [31]

Yes you were correct the answer was c

4 0

3 years ago

Read 2 more answers

A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a

Bezzdna [24]

Answer:

25 psi

Explanation:

The weight of the car is:

W = mg

W = 1550 kg * 9.8 m/s²

W = 15,190 N

Divided by 4 tires, each tire supports:

F = W/4

F = 15,190 N / 4

F = 3797.5 N

Pressure is force divided by area, so:

P = F / A

P = (3797.5 N) / (0.16 m × 0.14 m)

P ≈ 170,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 170,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 25 psi

8 0

3 years ago

Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitud

navik [9.2K]

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, $q_1=-6.29\times 10^{-6}\ C$

Second charged particle, $q_2=5.23\times 10^{-6}\ C$

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}$

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

4 0

3 years ago