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agasfer [191]
3 years ago
8

Which of the following would NOT be a medium that mechanical waves move through

Physics
1 answer:
MaRussiya [10]3 years ago
3 0

Answer:

i don't have any answer without the options

Explanation:

Where is the pic

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A team of bicyclists are on bikes that require 512 J of work to ride. It takes 432 J of work for the bike to turn the gears. Wha
Alja [10]

Answer:

84.4 %

Explanation:

Mechanical efficiency = output work/input work × 100 %

output work = 432 J of work for the bike to turn the gears

input work = 512 J of work to ride.

Mechanical efficiency =  432 J/512 J × 100 %

= 0.844 × 100%

= 84.4 %

8 0
3 years ago
WILL GIVE BRAINLY!!!
Gnoma [55]
The answer is apogee my dude
7 0
3 years ago
A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true o
madam [21]

Answer:

D. The motion cannot be determined without knowing the speeds of the objects before the collision.

Explanation:

This question is tricky! We know the object moving to the left has a greater mass than the one moving to the right. We'd <em>assume</em> they would move to the left because the leftwards object has a greater mass, right?

Not. So. Fast.

We can solve for the objects' final velocity using the formula for momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v .

Now here's where the trap is sprung: <em>we don't think about the equation</em>. This shows that the final velocity of the objects and the direction depends on both the mass of the objects <em>and</em> their initial velocity.

Basically, what if the 3 kg object is moving at 1 m/s and the 4 kg object is moving at –0.5 m/s? The objects would move to the <em>right</em> after the collision!

Do we know the velocity of these objects? No, right?

That means we <em>can't</em> determine the direction of their motion <u>unless we know their initial, pre-collision velocity</u>. This question is tricky because we look at the 4 kg vs. 3 kg and automatically assume the 4 kg object would dictate the direction of motion. That's not true. It depends on velocity as well.

I hope this helps you! Have a great day!

4 0
3 years ago
Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou
Gennadij [26K]

Answer:

a. The station is rotating at 1.496 \frac{rev}{min}

b. the rotation needed is 2.8502 \frac{rev}{min}

Explanation:

We know that the centripetal acceleration is

a_{c}= \omega ^2 r

where \omega is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

<h3>a. </h3>

The rotational speed  is :

2.7 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.02454 \frac{rad^2}{s^2} }

\omega  = 0.1567 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.1567 \frac{rad}{s}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 1.496 \frac{rev}{min}

<h3>b. </h3>

The rotational speed needed is :

9.8 \frac{m}{s^2} = \omega ^2 * 110  \ m

\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}

\omega  = \sqrt{ 0.08909 \frac{rad^2}{s^2} }

\omega  = 0.2985 \frac{rad}{s}

Knowing that there are 2\pi \ rad in a revolution and 60 seconds in a minute.

\omega  = 0.2985 \frac{rev}{min}  \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}

\omega  = 2.8502 \frac{rev}{min}

3 0
3 years ago
Read 2 more answers
PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST
stiv31 [10]

Explanation:

False

electron and proton attract each other

3 0
3 years ago
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