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lord [1]
3 years ago
11

3. Magnitude of a physical quantity is the number of times a standard quantity is present in it.

Physics
1 answer:
Nady [450]3 years ago
4 0
The number of times a standard quantity is present in the given physical quantity is called magnitude of a physical quantity
You might be interested in
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Nostrana [21]
I think c I’m not sure tho
6 0
2 years ago
How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?​
miss Akunina [59]

Answer:

The answer to your question is    Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

           Pe = mgh

Substitution

           Pe = (50)(5)(9.81)

Simplification

           Pe = 2452.5 J

8 0
3 years ago
Scientists study how the continents move. Why might scientists use a model
tigry1 [53]

Answer:

B. It is too slow to observe directly

Explanation:

They move too slow to be able to observe how they move.


I hope it helps! Have a great day!
bren~

3 0
2 years ago
What conversion factor is used to convert 20 cm to m
Anastaziya [24]
The conversion factor you use is 100 cm = 1 m.
You can divide 20 by 100 to get the answer.
20 cm/100 cm =.2 m
Hope this helped!
7 0
3 years ago
In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones
wel

Answer:

0.37 m

Explanation:

Let the shoulder be the origin.

The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.

The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones  from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m

Substituting these into the equation for the center of mass, we have

y = (m₁y₁ + m₂y₂)/(m₁ + m₂)

y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)

y = (1.20 kgm + 0.65 kgm)/5.0 kg

y = 1.85 kgm/5.0 kg

y =  0.37 m

The distance of the center of mass from the shoulder is thus y = 0.37 m

7 0
3 years ago
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