What 3 places on your body do not grow hair

If you are asked to modify the robots or drones that are currently used , what kind of modifications you would do and suggest an

Eddi Din [679]

Explanation:

In recent times of pandemic, robots can be use as replacement of labor in the industries. Mundane tasks can be programmed in their system so that they can used readily.

Drones can used delivery for essential goods and services, so that human interference can be least and the spread of virus can be curbed.

In a recent example, Argentina where aerial data has reportedly been used to accelerate the construction of emergency hospitals.

4 0

4 years ago

A football player kicks the ball at a 45 degree angle. Without an effect from the wind, the ball would travel 60.0m horizontally

Ronch [10]

B when the ball is at its maxium height

5 0

3 years ago

Please help, I do not understand

Anettt [7]

I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

   (a fraction of one orbit) = (the same fraction of the other orbit)
AND
   the total elapsed time is a common multiple of their periods.

Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds) = (K+1) (π/ω seconds)

   2Kπ/ω   =    Kπ/ω + π/ω

Subtract Kπ/ω :    Kπ/ω = π/ω

Multiply by ω/π :    K = 1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

   In the time it takes the slower planet to revolve once,
   the faster planet revolves twice, and catches up with it.

   It will be 2π/ω seconds before the planets line up again.

   When they do, they are again in the same position as shown
   in the drawing.

To describe it another way . . .

   When Kanye has completed its first revolution ...

   Bieber has made it halfway around.

   Bieber is crawling the rest of the way to the starting point while ...

   Kanye is doing another complete revolution.

   Kanye laps Bieber just as they both reach the starting point ...

   Bieber for the first time, Kanye for the second time.

You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.

5 0

3 years ago

Potential energy is the energy possessed by an object due to its ___ and kinetic energy is the energy an object has a result of

Nata [24]

1)due to it's position in the gravitational field
2)due to it's motion

6 0

3 years ago

Read 2 more answers

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$