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3 years ago

Why do reflecting telescopes usually have a secondary mirror in addition to a primary mirror?

1 answer:

8 0

Once the starlight or moonlight comes down the tube and hits the big main mirror at the bottom, it's reflected, and heads back up the tube, along the same way it came in. But if you're looking through an eyepiece, the eyepiece is on the SIDE of the tube. So you need another small mirror, to catch the focused light before it goes up out of the tube, and send it sideways to the eyepiece.

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A 50 kg child runs off a dock at 2.0 ms (horizontally) and lands in a waiting rowboat of mass 150 kg. At what speed does the row

Alex73 [517]

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

5 0

3 years ago

A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:

Ilia_Sergeevich [38]

Answer:

6.198 m/s²
4.48 s
453.77 m

Explanation:

5 0

3 years ago

A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is

Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

p₀ = m v

final attempt. after separation

$p_{f}$ = m /2 0 + m /2 v_{f}

p₀ = p_{f}

m v = m /2 $v_{f}$

v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

K₀ = ½ m v²

final energy

$K_{f}$ = ½ m/2 0 + ½ m/2 v_{f}²

K_{f} = ¼ m (2v)²

K_{f} = m v²

the expression for work is

W = ΔK = K_{f} - K₀

W = m v² - ½ m v²

W = ½ m v²

5 0

3 years ago

A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0

nikklg [1K]

Answer:

(a) 10 s

(b) 30 m/s

Explanation:

The motorist's position at time t is:

x = 15t

The officer's position at time t is:

x = ½ (3) t² = 1.5 t²

(a) When they have the same position, the time is:

15t = 1.5 t²

t = 0 or 10 s

(b) The officer's speed is:

v = 3t

v = 30 m/s

x = 15t = 150 m

6 0

3 years ago

A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

nignag [31]

Answer:

Explanation:

Given

mass of box $m=80 kg$

coefficient of kinetic friction $\mu _k=0.15$

coefficient of Static friction $\mu _s=0.30$

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

$f_r=ma$

where $f_r=frictional\ Force$

$a=acceleration$

$a=0$

$f_r=0$

4 0

3 years ago