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4 years ago

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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

1 answer:

S_A_V [24]4 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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What metric unit would you use to estimate the actual distance between Boston and New York?

mrs_skeptik [129]

I would use miles Because miles will give you less of an answer

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4 years ago

Adding or removing thermal energy to or from a substance does not always cause its temperature

julia-pushkina [17]

Change in thermal energy not always cause it's temperature change. It is the situation when water reaches either at 0 C or 100 C then thermal energy doesn't cause change in temperature instead it changes the state of matter.

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3 years ago

Read 2 more answers

determine the maximum angle theta for which the light rays incident on the end of the optical fiber of radius 1 mm are subhect t

Vesna [10]

Answer:

Explanation:

Let the critical angle be C .

sinC = 1 / μ where μ is index of refraction .

sinC = 1 /1.2

= .833

C = 56°

Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )

sin i / sinr = 1.2 , i is angle of incidence

sini = 1.2 x sinr = 1.2 x sin 34 = .67

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7 0

3 years ago

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object

tatiyna

Answer:

$D_T=18.567m$

Explanation:

From the question we are told that:

Acceleration $a=8.0 m/s^2$

Displacement $d=1.05 m$

Initial time $t_1=6.0s$

Final Time $t_2=2.5s$

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

$V^2=2as$

$V=\sqrt{2*6*1.05}$

$V=4.1m/s$

Generally the equation for Distance traveled before stop is mathematically given by

$d_2=v*t_1$

$d_2=3.098*4$

$d_2=12.392$

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity $v_3=0 m/s$

Initial velocity $u_3=4.1 m/s$

Therefore

Using Newton's Law of Motion

$-a_3=(4.1)/(2.5)$

$-a_3=1.64m/s^2$

Giving

$v_3^2=u^2-2ad_3$

Therefore

$d_3=\frac{u_3^2}{2ad_3}$

$d_3=\frac{4.1^2}{2*1.64}$

$d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

$D_T=d_1+d_2+d_3$

$D_T=5.125m+12.392+1.05 m$

$D_T=18.567m$

6 0

3 years ago

krek1111 [17]

Answer:

c

Explanation:

a vector quantity has both magnitude and direction

6 0

3 years ago