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3 years ago

To have momentum, an object must be ______. A. heavy B. massive C. moving D. accelerating

Physics

2 answers:

swat323 years ago

6 0

To have momentum an object must be C. Moving

Amiraneli [1.4K]3 years ago

3 0

Momentum of the object can be calculated by multiplying the mass of the object and the velocity of the moving object. In this case, the starting situation should be the object should be moving, else there is no velocity and thus momentum is equal to zero .Answer is C

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The moon is in a nearly circular orbit of radius r = 384000000 meters

PSYCHO15rus [73]

Answer:

$2.72\cdot 10^{-3} m/s^2$

Explanation:

The centripetal acceleration of an object in circular motion is the acceleration with which the object is attracted towards the center of the circular orbit. Mathematically, it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the orbit

The speed of the object is also given by the ratio between the circumference of the orbit and the orbital period, T:

$v=\frac{2\pi r}{T}$

Substituting into the previous equation, we find a new expression for the centripetal acceleration:

$a=\frac{4\pi^2 r}{T^2}$

In this problem:

- The radius of the orbit of the Moon is

$r = 384000000 m = 3.84\cdot 10^8 m$

- The period of the orbit is

$T=27.32 d \cdot 24\cdot 60\cdot 60 =2.36\cdot 10^6 s$

Therefore, the centripetal acceleration is:

$a=\frac{4\pi^2 (3.84\cdot 10^8)}{(2.36\cdot 10^6)^2}=2.72\cdot 10^{-3} m/s^2$

5 0

3 years ago

What is the Ramsar convention?

Mashutka [201]

The Ramsar Convention on Wetlands of International Importance Especially as Waterfowl Habitat is an international treaty for the conservation and sustainable use of wetlands.

5 0

3 years ago

A bus initially at rest accelerated of 4m/S2 at the end of 10 second find average velocity

elixir [45]

Answer:

V = 20m/s

Explanation:

Given the following data;

Acceleration = 4m/s²

Time = 10 secs

Initial velocity = 0m/s (since it's at rest).

To find the average velocity, we would use the first equation of motion;

$V = U + at$

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the equation, we have;

V = 0 + 4*10

V = 40m/s

To find the average velocity;

Average velocity = (U + V)/2

Average velocity = (0 + 40)/2

Average velocity = 40/2

Average velocity = 20m/s

6 0

3 years ago

A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng

frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

$Q_{1} =\frac{W}{n}$

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

$Q_{1} =\frac{750}{0.35} =2142.86kJ$

$Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ$

c) The efficiency of Carnot engine is:

$n=1-\frac{300K}{550K} =0.455$

The heat transfer to the heat engine is:

$Q_{1c} =\frac{750}{0.455} =1648.35kJ$

The heat transfer from the heat engine is:

$Q_{2c} =1648.35-750=898.35kJ$

4 0

3 years ago

An ideal gas is brought through an isothermal compression process. The 3.00 mol 3.00 mol of gas goes from an initial volume of 2

ozzi

Answer:

The answers can be found by considering the isothermal expansion equation as well as the ideal gas equation from where we have

The temperature T = 602.64K and the final pressure P = 110.24MPa

Explanation:

Numbeer of moles of gas = 3.00 mol

initial volume = 230.8×10−6 m3

final volume = 133.4×10−6 m3 .

released energy = 8240 J

Temperature = Constant = T

Pressure = $p_{f}$ =unknown

From the relation the combined ideal gas law, PV = nRT

Where R = 8.314 4621.JK−1mol−1

we have The release energy from compression P1V1 -P2V2

-qrev = -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 8240J

n = 3

Hence -nRTln( $\frac{V_{2} }{V_{1} }$ ) = 3×8.314 462×ln $(\frac{133.4}{230.8})$ × T= -8240 J

or -13.67×T = -8240J, thus T = -8240/-13.67 = 602.64K

The Final pressure is given by

PV = n×R×T from where we have V = final volume thus

P = (n×R×T)/V = (3×8.134×602.64)÷(133.4× $10^{-6}$ ) = 110237041.1 N/ $m^{2}$ = 110.237MPa

8 0

3 years ago