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Anna11 [10]
3 years ago
12

Item 16 The weight w (in pounds) of an object varies inversely with the square of the distance d (in miles) of the object from t

he center of Earth. At sea level (3978 miles from the center of Earth), an astronaut weighs 210 pounds. How much does the astronaut weigh 200 miles above sea level
Physics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

190.3759 pounds

Explanation:

If the astronaut weighs 210 pounds when he is at a distance of 3978 miles from the Earth's center, then he will weigh

210 [3978 / (3978 + 200) ]² = 190.3759 pounds

at an altitude of 200 km.

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Explanation:

Gravitational potential energy = mgh = (5)(9.81)(7) = 343.35J.

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The graph represents velocity over time.<br> What is the acceleration?
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Explanation:

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A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn
kow [346]

Answer:

The tension in the cord is T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} } (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

Mg=(M+\frac{MR^{2} }{2b^{2} }  )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2}  }

Replacing in eq. 1

T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

3 0
3 years ago
Read 2 more answers
Suppose you had two magnets, who were attracted to each other, and you brought them together. Then took them apart by separating
IgorLugansk [536]

Answer:4. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is tripled, then what is the new force? Explanation: The electrostatic force is inversely related to the square of the separation distance.

Explanation:

8 0
3 years ago
a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s
Pepsi [2]

Answer:

\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a:  \\  \sf \implies 0 = 30 + ( - 1.5)(t) \\   \sf  \implies 0 = 30 - 1.5(t) \\  \sf \implies 30 - 1.5(t) = 0 \\  \\  \sf Subtract  \: 30  \: from  \: both  \: sides: \\  \sf \implies (30 -  \boxed{ \sf 30}) - 1.5(t) =  \boxed{ \sf  - 30} \\  \\  \sf 30 - 30 = 0 :  \\  \sf \implies  - 1.5(t) =  - 30 \\  \\  \sf Divide  \: both  \: sides \:  of \:  - 1.5(t) =  - 30 \: by \:  - 1.5 :  \\  \sf \implies  \frac{  - 1.5(t)}{ \boxed{ \sf - 1.5}}  =  \frac{ - 30}{ \boxed{ \sf -1.5 }}  \\  \\  \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}}  = 1 :  \\  \sf \implies t =  \frac{ - 30}{ - 1.5}  \\  \\   \sf  \frac{ - 30}{ - 1.5}  =  \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}}  = 20 :  \\  \sf  \implies t = 20 \: sec

So,

Time in which train will come to rest = 20 seconds

4 0
3 years ago
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