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2 years ago

Determine the angular velocity ω of the telescope as it orbits around the Sun.

Physics

1 answer:

lara31 [8.8K]2 years ago

5 0

The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun

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Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am 1 m away from B , the path difference will be

8 - 1 = 7 m which is odd multiple of 1 or λ /2

So path difference becomes odd multiple of λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

8 0

3 years ago

Read 2 more answers

Problem 2.26 MasteringPhysics 10 of 16 Problem 2.26 When striking, the pike, a predatory fish, can accelerate from rest to a spe

cluponka [151]

final velocity = initial velocity + (acceleration x time) 
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration

distance = (initial velocity x time) + 1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
distance = 0.18 m

6 0

3 years ago

Read 2 more answers

A ball is projected vertically upward from the surface of the Earth with an initial speed of +49 meters per second. The ball rea

fenix001 [56]

Answer: 245 meters

Explanation: 49 ? Meters
over over over
1 5 seconds

Cross multiply- 49 times 5 divided by 1 which equals 245

8 0

2 years ago

Help plzzz!!! I only have 10 minutes to turn in

Lostsunrise [7]

The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

Kinetic energy = (1/2) (mass) (speed²)

And there we have it

The car's mass is 3,600 kg.
Its speed is 'v' m/s .
(1/2) (mass) (v²) = 5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

v² = (5406/1800) (Newton-meter/kg)

v² = (5406/1800) (kg-m²/s² / kg)

v² = (5406/1800) (m²/s²)

= = = = =

v = √(5406/1800) m/s

v = 1.733 m/s

4 0

3 years ago

A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu

Len [333]

Answer:

0.4 m/s²

Explanation:

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

Mass (m) = 1 kg

Radius (r) = 0.2 m

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Tangential acceleration (aₜ) =?

Next, we shall determine the angular acceleration (a) of the sphere. This can be obtained as follow:

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20/10

a = 2 rad/s²

Finally, we shall determine the tangential acceleration (aₜ) of the sphere. This can be obtained as follow:

The tangential acceleration (aₜ) and the angular acceleration (a) are related according to the equation:

Tangential acceleration (aₜ) = Angular acceleration (a) × Radius (r)

aₜ = ar

With the above formula, we can obtain the tangential acceleration (aₜ) as follow:

Radius (r) = 0.2 m

Angular acceleration (a) = 2 rad/s²

Tangential acceleration (aₜ) =?

aₜ = ar

aₜ = 2 × 0.2

aₜ = 0.4 m/s²

Therefore, the tangential acceleration is 0.4 m/s²

6 0

3 years ago