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2 years ago

7

_____ are both ways of preserving current energy resources.

1 answer:

MakcuM [25]2 years ago

3 0

Answer:

4

Explanation:

I think if not. I am sorry

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Write some interesting facts about isaac newton

Hatshy [7]

Sir Isaac Newton was an English Mathematician,physicist,astronomer,theologian and author who is widely recognized as one of the most influential scientist of all time because of his discovery of gravity(force that attracts a body toward center of earth).

8 0

2 years ago

In which example would the most transfer of energy take place from Mechanical energy to Thermal (heat) energy?

drek231 [11]

Answer:The answer is when a metal plate is pressed against a fast spinning piece of steel hard enough to stop it. i know this because I just took a test and got this question and got it write

Explanation:

5 0

2 years ago

In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col

saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ( $a$ ), measured in meters per square second, is estimated by this kinematic formula:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }$ (1)

Where:

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ , $v$ - Initial and final speeds of the spaceship, measured in meters.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v = 10968\,\frac{m}{s}$ and $\Delta s = 212.8\,m$ , then the acceleration experimented by the spaceship is:

$a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}$

$a = 282652.782\,\frac{m}{s^{2}}$

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0

2 years ago

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

3 0

2 years ago

A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th

defon

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

8 0

2 years ago