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3 years ago

Which is colder, 0°C or 20°F?

Physics

2 answers:

Allushta [10]3 years ago

8 0

20 degrees F because 0 degrees celcius is around 32 degrees F

Mashutka [201]3 years ago

7 0

In order to compare the two temperatures, we need to convert 20°F into Celsius. The formula we need to use is:

$T(^{\circ}C)=\frac{5}{9}(T(^{\circ}F)-32)$

Substituting 20°F, we find

$T(^{\circ}C)=\frac{5}{9}(20^{\circ}F-32) =-6.7^{\circ}C$

And this is less than 0°C, so the answer is

20°F is colder than 0°C.

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

A constant net force acts on an object. which of the following best describes the object's motion?

Anton [14]

The object<span> is moving with a decreasing acceleration. The </span>object<span> is moving with </span>a constant<span> velocity.</span>

4 0

3 years ago

A driver must always stop within 50 ft but not less than ____________ ft from the nearest rail when the signal is flashing and t

k0ka [10]

Answer:

20 ft

Explanation:

5 0

3 years ago

Read 2 more answers

The work function for magnesium is 3.70 ev. what is its cutoff frequency?

alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is 3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

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7 0

2 years ago

What type of change accurs when a substance stays the same

Thepotemich [5.8K]

This would be a physical change because it can change back to its original form. This is like ripping paper. You can piece it back together and it still is paper.

The opposite of this is chemical change. Chemical change means the product has been changed completely like burning paper. The paper has now been turned to ash and it's impossible to change this back to its original form.

3 0

3 years ago