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balu736 [363]
3 years ago
13

A concave mirror forms a real image at 25 cm from the mirror surface along the principal axis. If the corresponding object is at

10.0 cm distance, what’s the mirror’s focal length?

Physics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

7.143cm

Explanation:

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a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second
Gemiola [76]

1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.

The frequency of the radio station is:
f
=
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k
H
z
=
 1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W

The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s

The number of photon emitted per second = N = P/hf = <u>30.16 x 10^30 </u>photon/s.

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6 0
1 year ago
Define fission and give an example of a reaction
erastova [34]
<span>Definition: The action of dividing or splitting something into two or more parts.

Reaction Example: </span><span>Autotomy, sometimes termed transverse </span>fission<span>, is the name given to a process of unequal </span>fission<span> in which a portion of the body separates off with subsequent regeneration.</span><span>

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6 0
3 years ago
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Why did the following change occur? the snow that covered the ground all winter began to melt, temperatures rose, and flowers st
FromTheMoon [43]

Answer:C

Explanation:

8 0
2 years ago
Read 2 more answers
A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef
AveGali [126]
The thermal efficiency of an engine is
\eta= \frac{W}{Q}
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%
8 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
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