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bezimeni [28]
3 years ago
8

Rohith has been jogging to the bus stop for 2.0 minutes at 3.5 m/s when he looks at

Physics
1 answer:
nikdorinn [45]3 years ago
6 0
Multiply these numbers and there’s your answer
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A small rocket to gather weather data is launched straight up. Several seconds into the flight, its velocity is 140 m/s and it i
Free_Kalibri [48]

Explanation:

The force will be 98 N....

7 0
3 years ago
A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 2.9 m/s when it re
Rufina [12.5K]
This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
E_m=E_k+E_p
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
E_m=Ek=\frac{1}{2}mv^2
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
E_m=E_p
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
E_M=\frac{1}{2}mv^2
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
E_p=mgh=E_m=\frac{1}{2}mv^2
Solving for h gives us:
h=\frac{v^2}{2g}.
It doesn't depend on mass!

4 0
3 years ago
Read 2 more answers
PLEASE HELP!!!<br>How are middle latitude cyclones and tropical cyclones different from each other?
defon
In contrast, extratropical cyclones have their strongest winds near the tropopause, which is about 8 miles above the surface. These differences are due to the tropical cyclone being “warm-core” in the troposphere, whereas extra-tropical cyclones are “warm-core” in the stratosphere and “cold-core” in the troposphere.
8 0
3 years ago
A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?
kramer
To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
7 0
3 years ago
Answers to electrical?
Orlov [11]
Can you get a better pic so i can read it

3 0
3 years ago
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