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3 years ago

Which transformation of energy occurs when a fan is turned on and the blades start to rotate?

Physics

1 answer:

Ksju [112]3 years ago

5 0

Answer:

kinetic energy

Explanation:

A wind turbine transforms the mechanical energy of wind into electrical energy. A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator

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A railroad car of mass 2.00 3 104 kg moving at 3.00 m/s collides and couples with two coupled railroad cars, each of the same ma

FrozenT [24]

Given:

Mass of the rail road car, m = 2 kg

velocity of the three cars coupled system, v' = 1.20 m/s

velocity of first car, $v_{a}$ = 3 m/s

Solution:

a) Momentum of a body of mass 'm' and velocity 'v' is given by:

p = mv

Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:

$mv_{a} + 2mv_{b} = (m + 2m)v'$ (1)

where,

$v_{a}$ = velocity of the first car

$v_{b}$ = velocity of the 2 coupled cars after collision

Now, from eqn (1)

$v' = \frac{v_{a} + 2v_{b}}{3}$

$v' = \frac{3.00 + 2\times 1.20}}{3}$

v' = 1.80 m/s

Therefore, the velocity of the combined car system after collision is 1.80 m/s

7 0

3 years ago

Read 2 more answers

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

7 0

3 years ago

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend

Rasek [7]

Answer:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

3 0

3 years ago

How does increasing the distance between charged objects affect the electric force between them?

mina [271]

Answer: energy force

Explanation:

6 0

3 years ago

A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou

Sophie [7]

Answer:

442.36038 m or 1451.31362 ft

Explanation:

$P_1$ = Initial pressure = 30.15 inHg

$P_2$ = Final pressure = 28.607 inHg

$\rho$ = Density of air = 0.075 lb/ft³

$1\ lb/ft^3=16.0185\ kg/m^3$

$1\ in=0.0254\ m$

$1\ m=3.28084\ ft$

Density of mercury = 13560 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Difference in pressure is given by

$P_1-P_2=\rho gh\\\Rightarrow h=\frac{P_1-P_2}{\rho g}\\\Rightarrow h=\frac{(30.15-28.607)\times 13560\times 0.0254\times 9.81}{0.075\times 16.0185\times 9.81}\\\Rightarrow h=442.36038\ m\\\Rightarrow h=442.36038\times 3.28084\\\Rightarrow h=1451.31362\ ft$

The height of the building is 442.36038 m or 1451.31362 ft

4 0

3 years ago