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IgorLugansk [536]

2 years ago

11

The Enterprise goes into orbit around a mysterious planet. The ship moves at 4200 m/s in a circle of radius 4.91 x 10^7 m. What

is the mass of the
planet?

1 answer:

NeX [460]2 years ago

3 0

Answer:

M = 1.3*10^25 Kg

Explanation:

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A rock rolls down a hill. Which form of energy is this an example of? (2 points)

Bond [772]

Answer:

c.Mechanical

Explanation:

6 0

3 years ago

An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the

Korolek [52]

Efficiency = (useful output) / (input)

Efficiency = (35 J) / (125 J) = 0.28 = 28%

8 0

3 years ago

Read 2 more answers

ben walks 2 m from his desk to the teachers desk. From the teachers desk he then walks 3 m in the same direction to the classroo

vaieri [72.5K]

Distance is the total length covered = 2m + 3m = 5m

Displacement is his distance from original position.

Displacement = 2m + (-3)m. Representing the 3m walked back as -3.

Displacement = 2m - 3m = -1m.

So his displacement is 1m behind his original starting point.

4 0

2 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453

Drupady [299]

Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

$0.0001\times \frac{100}{0.4539}=0.022%$

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

$\frac{1}{0.022}\times {100}=4545.5\ kg$

Mass in pounds would be

$\frac{4545.5}{0.4539}=10014.32\ lb$

Mass in pound-mass is 10014.32 lb

8 0

3 years ago