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Trava [24]
3 years ago
6

Why is most space exploration accomplished with spacecraft that do not have astronauts on board

Physics
2 answers:
Cerrena [4.2K]3 years ago
7 0

Answer:

Explanation:

so people dont have to risk their lives to discover space. They want to test its safety out by using rovers and other robots.

Olin [163]3 years ago
6 0
Because of the enormous expense of creating conditions aboard
a spacecraft in which astronauts can survive, and the huge extra
weight and expense of guaranteeing their safety.
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while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w
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A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power
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Loud of dust and gas in space is a(n) Blank Space __________.
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A cloud in space that is composed of dust and gas is call a nebula. The word "nebula" is derived from the Latin word for "cloud," and nebulae in space are indeed large interstellar clouds made up of dust, hydrogen, helium and plasma.

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An isolated parallel-plate capacitor has a surface charge density. If the space between the plates is filled with a material of
anygoal [31]

Answer:

Explanation:

2\sqrt{34}

3 0
2 years ago
Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

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3 years ago
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