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Goshia [24]
3 years ago
15

What type of motion occurs when an object spends around and axis without altering its linear position?

Physics
1 answer:
ch4aika [34]3 years ago
7 0
I believe it is called centripetal force <span />
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How many minutes would it take a light wave to travel from the planet Venus to Earth? (Average distance from Venus to Earth = 28
Gnesinka [82]

Answer:

t=2.51min

Explanation:

The time taken by the light to travel a given distance is defined as:

t=\frac{d}{c}

Here c is obviously the speed of light. Now we convert the average distance form Venus to Earth to meters:

28*10^{6}mi*\frac{1609.34}{1mi}=4.51*10^{10}m

Finally, we calculate the minutes taken by the light to travel from Venus to Earth:

t=\frac{4.51*10^{10}m}{3*10^8\frac{m}{s}}\\t=150.33s*\frac{1min}{60s}=2.51min

6 0
4 years ago
The transfer of energy by the movement of particles that are in contact with each other
jeka57 [31]

Answer: I belive that the Answer is C.) Conduction

Explanation:

3 0
3 years ago
Read 2 more answers
A lemming running 2.22m/s runs off a horizontal cliff. It lands in the water 7.45m from the base of the cliff. How much time was
Gelneren [198K]
3,56.........................
6 0
4 years ago
If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre
Dmitrij [34]
M <span>represent mass in kg
</span><span>v represents speed in m/s
</span><span>r represents radius in m

Now, just substitute these into the formula:
</span>F =  \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }<span>

</span>
3 0
4 years ago
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi
Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

   P₁ = m v

   P₁ = 0.0575 x 26.7

   P₁ = 1.535 kg.m/s

b) momentum of ball after collision

   P₂ = m v'

   P₂ = 0.0575 x (-29.5)

   P₂ = -1.696 kg.m/s

c) change in momentum

    Δ P = P₂ - P₁

    Δ P = -1.696 -1.535

    Δ P = -3.231 kg.m/s

d) using conservation of momentum

  initial speed of racket = 0 m/s

  M u + m v = Mu' + m v

  M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

  0.331 u' = 3.232

     u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0
4 years ago
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