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Monica [59]
3 years ago
7

It was once recorded that a Jaguar

Physics
1 answer:
Artyom0805 [142]3 years ago
6 0

Answer:

71.85 m/s

Explanation:

Given the following :

Length of skid marks left by jaguar (s) = 290 m

Skidding Acceleration (a) = - 8.90m/s²

Final velocity of jaguar (v) = 0

Speed of Jaguar before it Began to skid =?

Hence, initial speed of jaguar could be obtained using the formula :

v² = u² + 2as

Where

v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar

0² = u² + (2 × (-8.90) × 290)

0 = u² + (-5,162)

u² = 5162

Take the square root of both sides

u = √5162

u = 71.847 m/s

u = 71.85m/s

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Just as a skydiver steps out of the helicopter with no forward velocity someone who’s watching start stopwatch so the time is ze
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Justification:


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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!!!

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Answer:

where is the box? ehh bring the box out

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