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Luden [163]
3 years ago
15

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Earth's pull?
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

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Superman takes 8 seconds to stop a runway train over a distance of 58 m using a power of
elena-14-01-66 [18.8K]

Answer:

Workdone = 600 Kilojoules

Explanation:

Given the following data:

Time = 8 seconds

Power = 75,000 Watts

Distance = 58 m

To find the work done;

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

Thus, work done is given by the formula;

Workdone = power * time

Workdone = 75000 * 8

Workdone = 600,000 = 600 KJ

5 0
3 years ago
Funnel shaped storm cloud formations are called
Olin [163]
A funnel shaped cloud is called a funnel cloud.
If it touches the ground it is called a tornado.
3 0
3 years ago
Read 2 more answers
Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation, in kW, from 1
Leya [2.2K]

Answer:

1.5 x 10⁵ W

Explanation:

A = Area of the fresh lava = 1.02 m²

T = Temperature of fresh lava = 1000 °C = 1000 + 273 = 1273 K

T₀ = Temperature of surrounding = 25.3 °C = 25.3 + 273 = 298.3 K

ε = emissivity of the lava = 0.97

σ = stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Rate of  transfer is given as

E = σ ε A (T⁴ - T₀⁴)

E = (5.67 x 10⁻⁸) (0.97) (1.02) ((1273)⁴ - (298.3)⁴)

E = 1.5 x 10⁵ W

3 0
3 years ago
A skydiver weighing 200 lbs with clothes that have a drag coefficient of .325 is falling in an area that has an atmospheric dens
TiliK225 [7]

Answer:

The right solution is:

(a) 89.455 m/s

(b) 44.73 m/s

Explanation:

The given values are:

Mass,

m = 200 lbs

or,

   = \frac{200}{2.205} \ kg

   = 90.7 \ kg

Air's density,

\delta = 1.225 \ kg/m^3

Drag coefficient,

C_d=0.325

When body is straight, area,

A_1=6 \ ft^2

As we know,

Terminal velocity,

⇒  V_t=\sqrt{\frac{2W}{C_d \delta A} }

or,

⇒      =\sqrt{\frac{2mg}{C_d \delta A} }

At straight orientation,

⇒ V_t'=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558} }

⇒      =\sqrt{\frac{1777.72}{0.223}}

⇒      =89.455 \ m/s

When belly flat,

⇒  V_t''=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558\times 4} }

⇒        =\sqrt{\frac{1777.72}{0.889} }

⇒        =44.73 \ m/s

3 0
3 years ago
The force required to open air bags is dangerous for:
pentagon [3]

It is dangerous for children. Air bags are dangerous to children age 12 and under because the bag inflates at speeds up to 200 mph and that sudden blast of energy can severely injure or kill passengers who are too close to the air bag. If possible, children should ride in the center of back seat, properly restrained with a seat belt.

5 0
3 years ago
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