4 years ago

Como responder Guess en physics

Physics

1 answer:

gizmo_the_mogwai [7]4 years ago

3 0

No entiendo la pregunta

You might be interested in

when astronauts work in space, they are sometimes attached to the spacecraft by a line called an umbilical. Why do you think the

wariber [46]

Probably for the umbilical cord that connects babies (from their early stages in the womb to their removal) to their mothers. The cord is cut, forming the belly button. This is analogous to astronauts in space.

3 0

4 years ago

Read 2 more answers

The rating of a light bulb at 100 watts indicates

Slav-nsk [51]

That marking means that when the expected voltage is connected
between the two contacts of the bulb, it will dissipate 100 joules of
energy every second, in the form of heat and light.

8 0

4 years ago

A 68.5kg astronaut floating motionless next to the space station throws a 2.25kg tool away from her at 3.20m/s. With what speed

valkas [14]

Answer:

-0.105 m/s

Explanation:

Given that

Mass of the astronaut, m(a) = 68.5 kg

Mass of the tool, m(t) = 2.25 kg

Speed of the tool after it is thrown, v(t) = 3.20 m/s

We know that momentum of a particle,

p = mv

See the attachment for calculations

Therefore, the speed is 0.105 m/s and it moves in the opposite direction.

5 0

4 years ago

Which of the materials would be able to scratch Quartz

Lelechka [254]

<span>anything harder than mohs scale 7 so eg Topaz, Corundum and diamond representing mohs scale 8 9 and 10 respectively.</span>

7 0

3 years ago

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular accelera

nalin [4]

Answer:

5.74s

Explanation:

We can first solve for the initial angular velocity using the following formula

$\omega^2 - \omega_0^2 = 2\alpha\theta$

Where $\omega = -22.4rad/s$ is the final angular velocity, $\alpha = -22.4 rad/s^2$ is the angular acceleration and $\theta = 0$ is the angular displacement

$22.4^2 - \omega_0^2 = 2*(-7.8)*0$

$\omega_0^2 = 22.4^2$

$\omega_0 = 22.4rad/s$

So for the wheel to get from 22.4 to -22.4 with angular acceleration of -7.8 then the time it takes must be

$t = \frac{\Delta \omega}{\alpha} = \frac{-22.4 - 22.4}{-7.8} = 5.74s$

8 0

3 years ago