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Oksanka [162]
3 years ago
5

Why does the surface of the sun appear to have a sharp edge?

Physics
1 answer:
Anna11 [10]3 years ago
6 0
The photosphere is relatively thin compared to the other atmospheric regions.
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Determine the angular speed, in rad/s of:
ryzh [129]

Answer:

Explanation:

A. The earth about its axis:

The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.

\frac{1 rev}{24hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s

B. The minute hand of the clock makes one revolution in 60 minutes

To convert this to rad per second, we have

\frac{1 rev}{60mins}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 1.745 \times10^-3rad/s

C. The hour hand of  a clock completes one revolution in 12 hours

\frac{1 rev}{12hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 1.454 \times10^-4 rad/s

D. an egg beater turning at 300 rpm.

\frac{300 rev}{1minute}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s=0.0218rad/s

6 0
3 years ago
A cat leaps to catch a bird. if the cat's jump was at 60.0° off the ground and its initial velocity was 2.74 m/s, what is the hi
user100 [1]

Answer:

0.29 m

Explanation:

7 0
3 years ago
In a nuclear reactor, each atom of uranium (of atomic mass 235 u) releases about 200 MeV when it fissions. What is the change in
Ket [755]

Answer:

0.002372187708 kg

Explanation:

Each atom of Uranium 235 releases 200 MeV = 200×10⁶×1.60218×10⁻¹⁹

= 200×1.60218×10⁻¹³ Joule

Number of atoms in a 2.6 kg sample mass = (2.6/0.235)×6.02214076×10²³

⇒Number of atoms in a 2.6 kg sample mass = 66.627×10²³ atoms

Change in energy = Change in mass / (speed of light)²

ΔE = Δmc²

⇒200×1.60218×10⁻¹³×66.627×10²³ = Δm×(3×10⁸)²

⇒Δm = 200×1.60218×10⁻¹³×66.627×10²³/(3×10⁸)²

⇒Δm = 2372.187708×10⁻⁶ kg

∴Change in mass = 0.002372187708 kg

7 0
3 years ago
a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground
Blababa [14]

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

5.01J = 1.53J + KE_2

KE_2 = 3.48J

3 0
3 years ago
A horse has a momentum of 25 kg*m/s and a velocity of 2.5 m/s. What is the horse mass?
Ne4ueva [31]
<h3>Answer:  10 kg</h3>

==================================================

Work Shown:

We have these variables

  • p = momentum = 25 kg*m/s
  • m = mass = unknown
  • v = velocity = 2.5 meters per second

Solving for the mass gets us...

p = m*v

25 = m*2.5

25/2.5 = m

10 = m

m = 10

The mass of the horse is 10 kg.

6 0
3 years ago
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