A 35 kg chair is lifted 5M off the ground. what is the potential energy?

One of the primary goals of the Kepler space telescope is to search for Earth-like planets. Data gathered by the telescope indic

amid [387]

Answer:

85.62 m

168.75 years

101.04 years

Explanation:

$L_0$ = Length of ship = 143 m

v = Velocity of ship = 0.8c

c = Speed of light

s = Distance to Boralis orbit = 135 ly

Gamma value

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \gamma=\dfrac{1}{\sqrt{1-\dfrac{0.8^2c^2}{c^2}}}\\\Rightarrow \gamma=1.67$

Length contraction is given by

$L=\dfrac{L_0}{\gamma}\\\Rightarrow L=\dfrac{143}{1.67}\\\Rightarrow L=85.62\ m$

The length is 85.62 m

Time taken

$t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{135}{0.8}\\\Rightarrow t=168.75\ years$

Time taken from the perspective one Earth is 168.75 years

Time dilation is given by

$t'=\dfrac{t}{\gamma}\\\Rightarrow t'=\dfrac{168.75}{1.67}\\\Rightarrow t'=101.04\ years$

The time taken from the perspective of the ship is 101.04 years

8 0

3 years ago

Using a rope that will snap if the tension in it exceeds 356 N, you need to lower a bundle of old roofing material weighing 478

klasskru [66]

Answer:

a) 2.5 m/s²

b) 6.12 m/s

Explanation:

Tension of rope = T = 356N

Weight of material = W = 478 N

Distance from the ground = s = 7.5 m

Acceleration due to gravity = g = 9.81 m/s²

Mass of material = m = 478/9.81 = 48.72

Final velocity before the bundle hits the ground = v

Initial velocity = u = 0

Acceleration experienced by the material when being lowered = a

a) W-T = ma

⇒478-356 = 48.72×a

$\Rightarrow \frac{122}{48.72} = a$

⇒a = 2.5 m/s²

∴ Acceleration achieved by the material is 2.5 m/s²

b) v²-u² = 2as

⇒v²-0 = 2×2.5×7.5

⇒v² = 37.5

⇒v = 6.12 m/s

∴ Velocity of the material before hitting the ground is 6.12 m/s

5 0

3 years ago

Why does the number of dwarf planets recognized by astronomers in the solar system sometimes increase?

gregori [183]

Because sometimes it happens that they discover a dwarf planet
that nobody ever knew about before. When that happens, they
ADD the new one to the list of known dwarf planets, and then the
total number of dwarf planets on the list increases by 1 .

4 0

3 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0

3 years ago

I need the solution to this

posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

$v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}$

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

$v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m$

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0

2 years ago