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Vaselesa [24]
3 years ago
14

The temperature of a system drops by 30°F during a cooling process. Express this drop in temperature in K, R, and °C.

Physics
1 answer:
babunello [35]3 years ago
8 0

Answer:

272.05 K; 489.69 °R; -1.11 °C

Explanation:

Fahrenheit can be converted to degrees Celsius using the following formula:

°C = 5/9 x (°F - 32) = -1.11 °C

The temperature in Kelvin is calculated from the temperature in degrees Celsius as follows:

K = °C + 273.15 = 272.05 K

The temperature on the Rankine scale is calculated from Kelvin as follows:

°R = 1.8K = 1.8(272.05) = 489.69 °R

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Acceleration is mathematically defined as; ΔV/t.

<h3>What is acceleration?</h3>

Acceleration is defined as the ratio of the change in velocity to time. It is defined as the extent to which the velocity changes within a given time interval.

The question is incomplete so we can not arrive at the final answer. However, acceleration is mathematically defined as; ΔV/t.

Learn more about acceleration: brainly.com/question/2437624

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2 years ago
When waves of equal amplitude from two sources are out of phase when they interact it is called?
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I believe destructive interference
6 0
3 years ago
A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate
maksim [4K]

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

        \texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

a=\frac{11.7-26.5}{5.2}=-2.85m/s^2

Angular acceleration

        \alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2

Angular acceleration = -3.95 rad/s²  

Magnitude =  3.95 rad/s²     

4 0
3 years ago
A ball is thrown straight upward. When it has reached the highest point in its motion, and is momentarily stopped, its accelerat
Damm [24]
Increased .....................
7 0
3 years ago
A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight
Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is  7.61 x 10⁻⁶  

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

  • m is mass supported by the steel
  • g is acceleration due to gravity
  • A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

  • E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

  • ΔL is change in length
  • L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0
1 year ago
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