Answer:
a) Team A will win.
b) The losing team will accelerate towards the middle line with 0.01 m/
Explanation:
Given that Team-A pulls with a force , 
and Team-B pulls with a force , 
∵ 
The rope will move in the direction of force 
.
∴ Team-A will win.
b) Considering both the teams as one system of total mass , 
Net force on the system , 
= 50-45 = 5N
Applying Newtons first law to the system ,
F = ma , where 'a' is the acceleration of the system.
Since , both the teams are connected by the same rope , their acceleration would be the same.
∴ 5 = 499×a
∴ a = 0.01 m/
Answer:
Atomic mass is defined as the number of protons and neutrons in an atom
Answer:
M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b]
ΔR is the difference in density between the gas inside and surrounding the balloon.
R[b] is the density of gas inside the baloon.
====================================
Let V be the volume of helium required.
Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V
U = 1.225gV newtons
----
Weight of Helium = Volume of Helium * Density of Helium * g
W[h] = 0.18gV N
Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N -----
Weight of 260kg = 2549.7 N
Then to lift the whole thing, F > 2549.7
So minimal F would be 2549.7
----
1.045gV = 2549.7
V = 248.8 m^3
Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)
=====
Let the density of the surroundings be R
Then U-W = (1-0.9)RgV = 0.1RgV
So 0.1RgV = 2549.7 N
V = 2549.7 / 0.1Rg
Assuming that R is again 1.255, V = 2071.7 m^3
Then mass of hot air required = 230.2 * 0.9R = 2340 kg
Notice from this that M = 2549.7/0.9Rg * 0.1R so
M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)
M[min] = M[basket] * ΔR/R[b]
Gas "floats" so if there are examples or pictures it would be the one with the most evenly spread out "dots".
Answer:
P = 0.0644 atm
Explanation:
Given that,
The pressure of a sample of gas is measured as 49 torr.
We need to convert this temperature to atmosphere.
The relation between torr and atmosphere is as follow :
1 atm = 760 torr
1 torr = (1/760) atm
49 torr = (49/760) atm
= 0.0644 atm
Hence, the presssure of the sample of gas is equal to 0.0644 atm.
