PLEASE HELP WILL GIVE MAX POINTS !!!!

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

noname [10]

Question:

The water molecules now in your body were once part of a molecular cloud. Only about onemillionth of the mass of a molecular cloud is in the form of water molecules, and the mass density of such a cloud is roughly 2.0×10−21 g/cm^3.

Estimate the volume of a piece of molecular cloud that has the same amount of water as your body.

Answer:

The volume of cloud that has the same density as the amount of water in our body is 1.4×10²⁵ cm³

Explanation:

Here, we have mass density of cloud = 2.0×10⁻²¹ g/cm^3

Density = Mass/Volume

Volume = Mass/Density = If the mass is 40 kg and the body is made up of 70% by mass of water, we have

28 kg water = 28000 g

Therefore the Volume = 28 kg/ 2.0×10⁻²¹ g/cm^3 = 1.4×10¹⁹ m³ = 1.4×10²⁵ cm³.

Therefore, the volume of cloud that has the same density as the amount of water in our body = 1.4×10²⁵ cm³.

6 0

3 years ago

A proton moves at a constant velocity of 50m/s along the x axis ,through crossed electric and magnetic fields .The magnetic fiel

shusha [124]

Answer:

42 I THE CORRECT ANSWER

Explanation:

BECAUSE I KNOW

6 0

3 years ago

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$

4 0

3 years ago

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 10

Tema [17]

Answer:

Explanation:

(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is 300m/150s=2ms^−1

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.

Therefore, his average speed for this journey is 400m210s=1.9ms−1.

For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.

Hence, his average velocity for this case is 200m/210s=0.95ms^−1

7 0

3 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m