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Leno4ka [110]
3 years ago
14

What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

and an average orbital speed of 29.7 km/s
Physics
1 answer:
masya89 [10]3 years ago
4 0

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

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An object is moving with uniform speed in a circle of radius r. Calculate the distance and displacement
dimaraw [331]

Answer and Explanation:

distance will be 2×3.14 (pie)×r

displacement will be 2r (diameter)

the motion is uniform circular motion as the object is moving in a circular path with uniform motion

8 0
2 years ago
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A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a
yawa3891 [41]
Write out what you have which is:
initial velocity 
final velocity 
Y distance 
degree

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X distance 
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from what you have you can plug into your formulas to get time.
6 0
3 years ago
Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in
sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, P_{atm} = 101 Kpa

P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}

P_{gauge}  = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

P_{gauge}  = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge}  = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge}  = 110,298.26 pa

Therefore, the gauge pressure of air is 110 kpa

4 0
4 years ago
Please help
JulsSmile [24]

Answer:

D

Explanation:

because it is the only one that has something to do with heat keyword would be boiling

7 0
3 years ago
A 16-slug mass is raised by 10 ft. the PE of the mass increased by?
Eva8 [605]

Answer:

<em>The PE of the mass increased by 6,972.95 J</em>

Explanation:

<u>Gravitational Potential Energy</u>

It's the energy stored in an object because of its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

We are given the mass of m=16 slug raised by a height h=10 ft. Both units will be converted to SI standard:

1 slug = 14.59 Kg, thus

16 slug = 16*14.59 Kg=233.44 Kg

1 ft = 0.3048 m, thus:

10 ft = 10*0.3048 m = 3.048 m

Thus, the PE of the mass increased by:

U = 233.44 * 9.8 * 3.048 = 6,972.95 J

the PE of the mass increased by 6,972.95 J

4 0
3 years ago
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