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Mamont248 [21]
2 years ago
5

You measure my life in hours and I serve you by expiring. I’m quick when I’m thin and slow when I’m fat. The wind is my enemy.

Physics
1 answer:
Pepsi [2]2 years ago
7 0
It’s a candle !! Mark me as brainliest please !!
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A 1.0 kilogram ball is thrown into the air with an initial velocity of 30m/s. How much kinetic energy does the ball have? (B) ho
Gala2k [10]

Answer:

450 joules ; 450 joules ; 45.9 m

Explanation:

Given that :

Initial Velocity, u = 30m/s

Mass, m = 1 kg

Kinetic Energy of ball (KE) = 0.5mu²

K. E = 0.5 * 1 * 30^2

K.E = 0.5 * 900

K.E = 450 Joules

B.) Potential Energy (P. E)

P. E = mgh

At the highest point, all kinetic energy has would have become potential energy, hence

K. E = P. E = 450 Joules

C) Height of the ball :

From ; P. E = mgh

Where ; g = acceleration due to gravity = 9.8m/s² ; h = height

450 = 1 * 9.8 * h

450 = 9.8h

h = 450 / 9.8

h = 45.918

h = 45.9 m

7 0
3 years ago
Why are weathering, erosion and deposition a NECESSARY process in the rock cycle?
Zarrin [17]

Answer:

YESS well it is partly nessary but it depends on the situation

Explanation:

4 0
2 years ago
Read 2 more answers
O professor Hosney, levou os alunos da segunda série ao laboratório para realizar um experimento. Pegou um recipiente de capacid
givi [52]

Answer:

The temperature beyond which the substance overflows the container is 86.23°C.

Explanation:

English Translation

Professor Hosney took the second grade students to the laboratory to perform an experiment. He took a 1000ml capacity container at a temperature of 68oF and poured 980 ml of a substance at 20oC into it. While placing the set to heat, he consulted a table where he found the volumetric expansion coefficient of the substance, 4 x 10-4 ºC-1 and the linear expansion coefficient of the container material, 3 x 10-5 ºC-1. Hosney then asked students to determine the temperature from which the substance would overflow. A student then asked, what is the melting temperature of the substance, and the teacher answered promptly 290.8 K. What is the temperature from which the substance will overflow?

Solution

The change in volume of a substance is given as

ΔV = γV₀(ΔT)

where

γ = coefficient of volume expansion

V₀ = Initial volume

(ΔT) = change in temperature.

At the temperature where the substance will overflow, the volume of the substance and the container will both be the same.

Let this temperature be T.

For the substance,

γ = coefficient of volume expansion = (4 × 10⁻⁴) °C⁻¹

V₀ = Initial volume = 980 mL

(ΔT) = change in temperature = (T - 20)

We will still leave ΔT as ΔT

ΔV₁ = (4 × 10⁻⁴) × 980 × ΔT

ΔV₁ = 0.392 ΔT

New volume of the substance at that temperature = V₀ + ΔV₁ = 980 + 0.392ΔT

For the container

γ = coefficient of volume expansion = 3 × coefficient of linear expansion = 3 × (3 × 10⁻⁵) °C⁻¹ = (9 × 10⁻⁵) °C⁻¹

V₀ = Initial volume = 1000 mL

(ΔT) = change in temperature = (T - 20) (note that 68°F = 20°C)

We will still leave ΔT as ΔT

ΔV₂ = (9 × 10⁻⁵) × 1000 × ΔT

ΔV₂ = 0.09 ΔT

New volume of the container at that temperature = V₀ + ΔV₂ = 1000 + 0.09 ΔT

At the temperature where overflow occurs, the two volumes are initially first the same.

980 + 0.392ΔT = 1000 + 0.09 ΔT

0.392ΔT - 0.09ΔT = 1000 - 980

0.302ΔT = 20

ΔT = (20/0.302) = 66.23°C

T - 20° = 66.23°

T = 66.23 + 20 = 86.23°C

The temperature beyond which the substance overflows the container is 86.23°C.

Hope this Helps!!!

8 0
3 years ago
Which of the following statements is not consistent with the properties of a molecular solid?a. a compound that conducts electri
erica [24]

Answer:

The correct answer to the question is

a. a compound that conducts electricity when molten

Explanation:

A molecular is solid comprises of a collection of molecules combined together by van der Waals forces and not having covalent or ionic bonds. They are made up of forces atoms or molecules combined by dipole-dipole forces, London dispersion forces or hydrogen bonds. Exmples include graphite and diamonds. The forces between the molecules in  molecular solid is also known as intermolecular forces (IMFs)

5 0
3 years ago
You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom flo
maksim [4K]

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

\text{Mass of the child} = m = 25kg

\text{Acceleration due to gravity} = g = 9.81m/s^2

\text{Height lifted} = h = 0.80m (Upward)

Work done to upward the object

W = mgh

W = (25)(9.81)(0.8)

W = 196.2J

Horizontal Force applied while carrying 10m,

F = 0N

W = 0J

Height descended in setting the child down

h' = -0.8m (Downwoard)

W = mgh'

W = (25)(9.81)(-0.80)

W = -196.2J

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

6 0
3 years ago
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