Answer:
For mass m 1 newton 2nd law
F=m 1 a 1
5=m 1 ×10
m 1 = 2
1
kg
For mass m 2
F=m 2 a 2
5=m 2
×20
m 2
= 4
1 kg
if tied together
Total mass =m 1 +m 2 = 1/2 +1/4=3/4kg
Now
F=M
T Q
T
a T = 5/m T = 5×4/3 = 3
/20m/s^2
Explanation:
Answer:
Volume will be 15 mL. Solution:- If we look at the given information then it is Boyle's law as the temperature is constant and the volume changes inversely as the pressure changes. So, the volume of the air bubble at the surface will be 15 mL.
Answer:
1. They both uses same energy
2. The 6 kg ball requires more power than 3kg ball
Explanation:
Sample 1
m = 3kg
g= 10m/s^2
h = 2m
t = 2secs
W = mgh = 3 x 10 x 2 = 60J
P= w/t = 60/2 = 30watts
Sample 2
m = 6kg
g= 10m/s^2
h = 1m
t = 1sec
W = mgh = 6 x 10 x 1 = 60J
P= w/t = 60/1 = 60watts
They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball
Answer:
6.75×10^13N
Explanation:
The electric force between the charges can be determined using coulombs law which states that 'the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of their distance between them.
Mathematically, F = kq1q2/r² where;
q1 and q2 are the charges
r is the distance between the charges
F is the force of attraction
k is the coulombs constant
Given q1 = 5C q2 = 15C r = 10cm = 0.1m k = 9×10^9Nm²/C²
Substituting the given values in the formula we have;
F =9×10^9×5×15/0.1²
F = 6.75×10^11/0.01
F = 6.75×10^13N
Therefore the electric force between them is 6.75×10^13N
Answer:


Explanation:
At initial condition
P=2 MPa
T=30°C
V=25 m/s
At final condition
P=0.3 MPa
Now from first law for open system

We know that for air
h= 1.010 x T KJ/kg
-----1
Now from mass balance

We also know that


----------2
Now from equation 1 and 2

So we can say that

This is the outlet velocity.
Now by putting the values in equation 2

This is the outlet temperature.