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suter [353]
4 years ago
6

What would happen if you drop and egg

Physics
2 answers:
Vaselesa [24]4 years ago
6 0
It would crack because of the pressure but it depends on the height of the drop
dexar [7]4 years ago
5 0

Answer:

It would crack.

Explanation: The pressure from dropping it would crush the eggshell therefore breaking the egg.

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As you can see, my cousin has a lot of hair. He uses an 1800 W blow dryer and it takes him
maw [93]

Power = 1800W (or 1.8KW by dividing by 1000)

Time = 3 hours

Power = energy/ time

1.8KW = energy/ 3

x3

5.4Kw/h= energy

(5.4KJ or 5400J used)

$0.15 Kw/h

$0.15 X 5.4 = 0.81

Thus, cost $0.81

Hope this helps!

5 0
2 years ago
(b) A piece of wood of volume 0.6 m² floats in water. Find the volume
enot [183]

Answer:

Explanation:

Let the volume below water be v . Then

buoyant force = v d g where d is density of water , g is acceleration due to gravity

= v x 1000 x g

weight of wood piece = volume x density of wood x g

= .6 x 600 x g

for equilibrium while floating

buoyant force = weight

= v x 1000 x g  =  .6 x 600 x g

v = .36 m²

volume above water or volume exposed = .6 - .36

= .24 m²

When immersed completely ,

buoyant force = .6 x 1000 x 9.8

= 5880 N

weight of wood

=  .6 x 600 x g

= 3528 N

buoyant force is more than the weight . In order to equalise them for floating with full volume in water

weight required = 5880 - 3528

= 2352 N.

6 0
3 years ago
The si base unit for length is the
Naya [18.7K]
The meter is the S.I.unit for length. 
8 0
3 years ago
What is harmonic oscillation ​
inn [45]

Explanation:

A simple harmonic oscillation is an oscillator that is neither driven nor damped.

3 0
3 years ago
Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the
Svet_ta [14]

Answer:

b)  x_{cm} = 4.88 cm , c) x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃) and d)

x_{cm}’= 1.88 cm

Explanation:

The definition of mass center is

    x_{cm} = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

    D₂ = d₁ + d₂

    D₂ = 1.1 + 1.9

    D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

    D₃ = D₂ + d₂

    D₃ = 3.0 + 3.9

    D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

    M = m₁ + m₂ + m₃

    M = 24 + 12 + 56

    M = 92 g

    x_{cm} = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

    x_{cm} = 1/92 (448.8)

    x_{cm} = 4,878 cm

    x_{cm} = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at   d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

    x_{cm} ’= 1/M  (m₁ d₁ + m₂ d₂ + m₃ d₃)

   

   x_{cm}’= 1/M  (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

    x_{cm}’= 1/92 ((24 (-1.9) +56 3.9)

    x_{cm}’= 1/92 (172.8)

    x_{cm}’= 1.88 cm

This distance is to the right of the central marble

3 0
3 years ago
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