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Semenov [28]
2 years ago
12

What is the formula that can be used to find velocity if kinetic energy and mass are known?

Physics
2 answers:
Temka [501]2 years ago
8 0

Answer:

v = √(KE*2/m)

Explanation:

The formula to compute kinetic energy (KE) is:

KE = 1/2*m*v^2

where m refers to mass and v refers to velocity. Isolating v we get:

KE = 1/2*m*v^2

KE*2/m = v^2

v = √(KE*2/m)

Given kinetic energy and mass, then velocity can be computed.

viva [34]2 years ago
6 0
The formula for kinetic energy is \frac{1}{2}m\Delta v^2. Thus, the equation for velocity is v=   \sqrt{ \frac{2TotalKineticEnergy}{m} }. 
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If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?
Vinvika [58]
The acceleration is0.25m/s^2
4 0
1 year ago
Read 2 more answers
Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
3 0
2 years ago
The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma
Molodets [167]

Answer:

the order of importance must be     b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

         b e a f c

5 0
2 years ago
Wood block 1 in (Figure 1), which has a mass of 1.0 kg, is at rest on a wood ramp. The angle of the ramp is 20º above horizontal
Sergeeva-Olga [200]

Answer:

Explanation:

For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.

X axis

      -Aₓ - f_e +T = 0        (1)

Y axis

      N₁ - W_y = 0              ( 2)

let's use trigonometry for the weight components

      sin θ = Wₓ / W

      cos θ = W_y / W

      Wₓ = W sin θ

      W_y = W cos θ

We write the diagram for the second body.

Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards

      W₂ -T = 0                             (3)

we add the equations is 1 and 3

       - W₁ sin θ - μ N₁ + W₂ = 0

from equation 2

       N₁ = W₁ cos θ

       

we substitute

        -W₁ sin θ - μ (W₁ cos θ) + W₂ = 0

W₂ = m₁ g (without ea - very expensive)

This is the smallest value that supports the equilibrium system

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2 years ago
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SpyIntel [72]

Answer:

uKkskdjod 7q and the rays are the best in all the ways ❤ ♥

7 0
2 years ago
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