All categories

3 years ago

Convert 34 kelvin to farenheit

Physics

1 answer:

Anuta_ua [19.1K]3 years ago

7 0

Answer:

-398.47

Explanation:

You might be interested in

Two rams run toward each other. One ram has a mass of 49 kg and runs west

olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram $m_1=49\ kg$

The velocity of this ram is $v_1=-7\ m/s$

Mass of the second ram $m_2=52\ kg$

The velocity of this ram $v_2=9\ m/s$

They combined after the collision

Conserving the momentum

$\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]$

Momentum after the collision will be

$\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}$

Therefore, option (d) is correct

4 0

3 years ago

How will decreasing the amplitude of the sound waves from a television affect its intensity?

Colt1911 [192]

A. The sound will decrease in volume

6 0

3 years ago

About how long does it take ocean water to travel around a gyre?

mylen [45]

The Gulf Stream, for example, has an average speed of 4 mph, or 100 miles/day, or 3600 miles/year, in round numbers. A complete circuit around the Atlantic is 8 or 10 thousand miles, which would take around 3 years on a raft.

4 0

4 years ago

Aliun [14]

Answer:

the correct answer is c, they will accelerate away from each other at different speeds. the 80kg will go faster due to less mass

6 0

3 years ago

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$

4 0

3 years ago