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Shtirlitz [24]
3 years ago
13

A skier, starting from rest, slides down an icy frictionless 80o incline whose vertical height is 105 m. How fast are they going

when the reach the bottom?
Physics
1 answer:
leonid [27]3 years ago
4 0

Answer:

nadaáaaaaaaaaaaaaa aaaaaaaaaaaaaaa

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4. How much time does it take for a student running at a speed of 5 m/s to cover a distance of 2,000 m?
Monica [59]

Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

Those who found this helpful please give me a Thanks to support me. So, I can explain other questions more clearly. If you don't want to mark me Brainliest don't mark. But, please give me a Thanks.

6 0
3 years ago
When looking through a fish tank, the mediums which light passes through are:
Alinara [238K]

The answer is : C ) air,water,and the tank glass.
-Hope this helps.

3 0
3 years ago
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A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done o
Vera_Pavlovna [14]

Answer:

0 J

Explanation:

As work is force times displacement, if no displacement occurs, no work occurs.

5 0
2 years ago
An example of forced convection is the movement of hot air by a fan.
zavuch27 [327]
This is true. let me know if im wrong.
4 0
3 years ago
A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77
Vikki [24]
The formula for the rotational kinetic energy is

KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2}  =2.07459 \ kgm^{2}

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}

KE_{rot} =18,671.31 \ J

6 0
3 years ago
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