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abruzzese [7]
3 years ago
9

What is the relationship between the weight of objects and the sum of their parts?

Physics
1 answer:
alexandr1967 [171]3 years ago
4 0

Answer:

No matter how parts of an object are assembled, the weight of the whole object is always the same as the sum of the parts; and when a thing is broken into parts, the parts have the same total weight as the original thing.

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As the moon revolves around the Earth, it also rotates on its axis. Why is it that the the same side of the moon is always visib
Nesterboy [21]
The moon dosent revolve around earth as the earth rotates the moon stays in the same place
7 0
3 years ago
A swimming pool is circular with a 30-ft diameter. The depth is constant along east-west lines and increases linearly from 5 ft
Schach [20]

Answer : The volume of water in the pool is, 2473 ft³

Explanation :

First we have to calculate the average depth.

Given:

Diameter = 30 ft

Depth range : 1 to 6 ft linearly

average depth =

Now we have to calculate the volume of water in the pool.

Volume = area × average depth

V = π × (radius)² × 3.5

V = π × (30/2)² × 3.5

V = π × (15)² × 3.5

V = π × 225 × 3.5

V = 3.14 × 225 × 3.5

V = 2472.75 ft³ ≈ 2473 ft³

Therefore, the volume of water in the pool is, 2473 ft³

3 0
4 years ago
Read 2 more answers
How does putting a mirror on a sharp bend help to prevent accidents
timofeeve [1]
It allows people to see around the corner and anticipate what speed changes they will need to make
6 0
3 years ago
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using hooke's law F spring = k Δ x find the force needed to stretch a spring 2 cm if it has elastic constant of 3 N/cm. A. 2/3 N
Assoli18 [71]
Data:
<span>Hooke represented mathematically his theory with the equation:
 
F = K * x
 
On what:
F (elastic force) = ?
K (elastic constant) = 3 N/cm
x (deformation or elongation of the elastic medium) = 2cm

Solving:

</span>F = K * x
F = 3 \frac{N}{\diagup\!\!\!\!\!\!cm}  * 2\diagup\!\!\!\!\!\!cm
\boxed{\boxed{F = 6N}}

Answer:
<span>C. 6 N</span>
4 0
4 years ago
Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re
Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0
3 years ago
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