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STatiana [176]
2 years ago
10

If the speed of an object in uniform circular motion is constant and the radial distance is doubled, by what factor does the mag

nitude of the radial acceleration decrease?
Physics
1 answer:
STatiana [176]2 years ago
6 0

clean and healthy

Explanation:

Because you will smell

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An integrated circuit is a ___.
EastWind [94]
HI buddy!

An integrated circuit(IC) is a semiconductor water on which thousands of tiny resistors, transistors are fabricated.

Remember that an integrated circuit are the heart and brains of most circuits. They are the keystone of modern electronics.

I hope this helps!
5 0
3 years ago
Investigations provide large amounts of information about a wide range of variables.
Klio2033 [76]

Hello this is to other people looking for the answer. Everyone else is wrong. I just took the quiz on e2020. The answer actually Comparative. Your welcome. Have a nice day.

8 0
3 years ago
What type of general wave requires a medium for the wave to travel through
Gemiola [76]

Answer:

Mechanical

Explanation:

Electromagnetic waves are waves that have no medium to travel whereas mechanical waves need a medium for its transmission.

7 0
2 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t
Bumek [7]

Explanation:

Since, it is mentioned the there occurs no change in the temperature. This also means that there will occur no change in thermal energy of the system.

Hence, \Delta E = 0. And, as \Delta E = 0 then there will be no work involved. This means that total energy added to the house will return to the outside air as heat.

Therefore,

                   Q = -(19000 J + 2000 J)

                       = -21000 J

or,    |Q| = 21000 J

Thus, we can conclude that the magnitude of the energy transfer between the house and the outside air is 21000 J.

5 0
3 years ago
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