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Digiron [165]
3 years ago
13

What is the weight force of a 10 kg object on the sun? (The sun's acceleration due to gravity is 275 m/s^2)?​

Physics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

f=mass × acceleration

f=10×275

f=2750N

Explanation:

PLS MARK BRAINLIEST

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I think y = 49x+3.5  

Explanation:

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What happens to the chemical bonds during energy exchange?
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A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
kap26 [50]

Answer:

a) \omega \approx 219.911\,\frac{rad}{s}, b) \alpha = 16.916\,\frac{rad}{s^{2}}, c) a_{t} = 1.776\,\frac{m}{s^{2}}, d) a_{n} = 5077.889\,\frac{m}{s^{2}}, e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) v = 23.091\,\frac{m}{s}.

Explanation:

a) The maximum angular velocity of the fan is:

\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

\omega \approx 219.911\,\frac{rad}{s}

b) The angular acceleration of the fan is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}

\alpha = 16.916\,\frac{rad}{s^{2}}

c) The magnitude of the tangential aceleration is:

a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)

a_{t} = 1.776\,\frac{m}{s^{2}}

d) The magnitude of the centripetal acceleration is:

a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)

a_{n} = 5077.889\,\frac{m}{s^{2}}

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)

v = 23.091\,\frac{m}{s}

5 0
3 years ago
A ship's boiler steam comes out at 112 C and pushes through the system, exiting into the condenser,
solniwko [45]

Answer:

D: 0.239

Explanation:

Equation for ideal efficiency is;

η = 1 - (T_c/T_h)

We are told that;

steam comes out at 112° C. Thus, T_h = 112°C. Converting to Kelvin gives; T_h = 112 + 273 = 385 K

The one exiting into the condenser is kept at 20°C. Thus; T_c = 20 + 273 = 293 K

Thus;

η = 1 - (293/385)

η = 0.239

5 0
3 years ago
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