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3 years ago

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Remember KE=M*V2/2…  A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy

of the slower airplane, the kinetic energy of the faster airplane is

1 answer:

pishuonlain [190]3 years ago

4 0

Answer: 4 times as much

Explanation:

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At Alpha Centauri's surface, the gravitational force between Alpha Centauri and a 2 kg mass of hot gas has a magnitude of 618.2

sergeinik [125]

Answer:

6.86 * 10^8 m

Explanation:

Parameters given:

Mass of hot gas, m = 2 kg

Gravitational Force, F = 618.2 N

Mass of Alpha Centauri, M = 2.178 * 10^30 kg

The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:

F = (G*M*m) / r²

Where G = gravitational constant

Therefore,

618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²

=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2

r² = 4.699 * 10^17 m²

=> r = 6.86 * 10^8 m

We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.

The mean radius of Alpha Centauri is 6.86 * 10^8 m.

5 0

3 years ago

Read 2 more answers

A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 59.0 lb/in2. If the pump is a cylinder of lengt

bagirrra123 [75]

Answer:

∴The air cannot be made to flow in with the given pump at the given conditions.

Explanation:

Given:

gauge pressure of bicycle tyre, $P_g=59\ lb.in^{-2}$

length of cylinder of the pump, $l=17.4\ in$

area of the the cylinder of the pump, $a= 3\ in^2$

we have the density of air at STP, $\rho=4.4256\times 10^{-5}\ lb.in^{-2}$

The piston must be pushed more than the pressure inside the tyre:

$P_g=\rho\times V\div a$

$59=4.4256\times 10^{-5}\times a\times h\div a$

$h=13.33\times 10^5\ in$

∴The air cannot be made to flow in with the given pump at the given conditions.

4 0

3 years ago

Difference betwwen Acclerat<br>ion and retradation

BartSMP [9]

I’m pretty sure one increases just straight up velocity and the other is kind of a deceleration

6 0

3 years ago

A college student is working on her physics homework in her dorm room. her room contains a total of 6.0×1026 gas molecules. as s

ella [17]

<span>6.6 degrees C Let's model the student as a 125 w furnace that's been operating for 11 minutes. So 125 w * 11 min = 125 kg*m^2/s^3 * 11 min * 60 s/min = 82500 kg*m^2/s^2 = 82500 Joule So the average kinetic energy increase of each gas molecule is 82500 J / 6.0x10^26 = 1.38x10^-22 J Now the equation that relates kinetic energy to temperature is: E = (3/2)Kb*Tk E = average kinetic energy of the gas particles Kb = Boltzmann constant (1.3806504Ă—10^-23 J/K) Tk = Kinetic temperature in Kelvins Notice the the energy level of the gas particles is linear with respect to temperature. So we don't care what the original temperature is, we just need to know by how much the average energy of the gas particles has increased by. So let's substitute the known values and solve for Tk E = (3/2)Kb*Tk 1.38x10^-22 J = (3/2)1.3806504Ă—10^-23 J/K * Tk 1.38x10^-22 J = 2.0709756x10^-23 J/K * Tk 6.64 K = Tk Rounding to 2 significant digits gives 6.6K. So the temperature in the room will increase by 6.6 degrees K or 6.6 degrees C, or 11.9 degrees F.</span>

7 0

3 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago