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malfutka [58]
3 years ago
6

1pt If the room is the frame of reference for the study of an object moving in three dimensions, from where should distances be

measured?
O A. one wall and the floor
OB. two walls and the floor
O c. the floor and the ceiling
O D. the door

Physics
1 answer:
oksian1 [2.3K]3 years ago
5 0

Answer:

Option B, two walls and the floor

Explanation:

The distance should be measured from the point where at least the three axed meet.

Two walls and the floor are equivalent to three axes.

Vertical wall 1 = Y Axis

Horizontal wall2 = X Axis

Floor = Z Axis

Thus, the distance should be measured from the point where two walls and one floor meet.

Option , B is correct

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At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp
motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

K = \frac{1}{2} mv^2

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

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3 years ago
Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w
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I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.
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Is the acceleration change or constnt?​
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Change.

Acceleration means going faster
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3 years ago
The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in
MissTica

Answer:

<em>155.80rad/s</em>

Explanation:

Using the equation of motion to find the angular acceleration:

\omega_f = \omega_i + \alpha t

\omega_f is the final angular velocity in rad/s

\omega_i  is the initial angular velocity in rad/s

\alpha is the angular acceleration

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Given the following

\omega_f = 6100rpm

Time = 4.1secs

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x  = 638.792rad/s

\omega_f = 638.792rad/s\\

Get the angular acceleration:

Recall that:

\omega_f = \omega_i + \alpha t

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

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3 years ago
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