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3 years ago

What is the average temperature of the city, 89.6 Fahrenheit, on the Celsius scale?

Physics

1 answer:

DaniilM [7]3 years ago

8 0

Answer:

thrity-two degrees Celsius

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Of all planets in our solar system jupiter has the greatest gravitational

sashaice [31]

<span>Of all planets in our solar system Jupiter has the greatest gravitational "Force as it is heaviest Planet in the solar system"

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5 0

3 years ago

Read 2 more answers

Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of

Veseljchak [2.6K]

As we know that as per Newton's II law we have

$F = \frac{dP}{dt}$

here we will have

$dP$ = change in momentum

$dt$ = time interval in which momentum is changed

now in order to have least injury during jumping we need to have least force on the jumper

so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least

So we need to increase the time in which momentum of the system is changed

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3 years ago

Snow and sleet when they fall to the ground is solar energy or gravitational force?

Nikolay [14]

Gravitational energy

3 0

3 years ago

on an unknown temperature scale, the freezing point of water is -15°U and the boiling point is +60°U. develop a linear conversio

shutvik [7]

Answer:

Since this is a linear equation

y = m x + b or

U = m F + b is a linear equation

when ΔF = (212 - 32) = 180

and ΔU = (60 - (-15)) = 75

m = 75 / 180 = 2.4 if converting F to U and a = .417

U = .417 F + b

If F = 32 then U = -15 and

-15 = .417 * 32 + b

b = -15 - 13.3 = -28.3 and our equation becomes

U = .417 F - 28.3

Check: let F = 212

U = .417 * 212 - 28.3 = 60 as it should

5 0

2 years ago

The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in

N76 [4]

Answer:

28.2 m/s

Explanation:

The range of a projectile launched from the ground is given by:

$d=\frac{v^2}{g}sin 2\theta$

where

v is the initial speed

g = 9.8 m/s^2 is the acceleration of gravity

$\theta$ is the angle at which the projectile is thrown

In this problem we have

d = 81.1 m is the range

$\theta=45^{\circ}$ is the angle

Solving for v, we find the speed of the projectile:

$v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s$

7 0

3 years ago