Answer:
8W
Explanation:
Given parameters:
Distance covered = 20m
Work done = 120J
Time = 15s
Unknown:
How much power does the Ox exerts = ?
Solution:
Power is the rate at which work is being done
Power = 
Power exerted by ox = 
= 8W
To determine the velocity in the restriction, we need to remember the law of conservation of mass where mass in should be equal to the mass out since mass cannot be created or destroyed. For this system, mass flow rate in is equal to mass flow rate out. We do as follows:
Mass flow rate in = Mass flow rate out
density (volumetric flow rate in) = density (volumetric flow rate out)
Since the liquid in and out are the same, then density would be cancelled.
(volumetric flow rate in) = (volumetric flow rate out)
Area in x velocity in = Area out x velocity out
velocity out = area in x velocity in / area out
velocity out = π (4/2)^2 (1.0) / π (2/2)^2
velocity out = 4 m/s
D. Adding another base pair will re-arrange your DNA sequences and cause an insertion mutation. This will make your codons group of differently, and possibly give you a BAD mutation. However, sometimes the codons still make the same proteins as its supposed to, the mutation will NOT affect you.
Example:
THE BIG FAT CAT ATE THE RAT
Now, if i were to add a letter (X) to this and make the letters group up in three aka the codons:
THE XBI GFA TCA TAT ETH ERA T
As you can you can see, adding a base pair in a DNA insertion will usually have a negative effect, specifically a insertion mutation.
Answer:
1.98 m/s
Explanation:
To solve this, we would be using the law of conservation of energy, i.e total initial energy is equal to total final energy.
E(i) = E(f)
mgh = ½Iw² + ½mv²
Recall, v = wr, thus, w = v/r
Also, I = ½mr²
I = 0.5 * 5 * 2²
I = 10 kgm²
Remember,
mgh = ½Iw² + ½mv²
Substituting w for v/r, we have
mgh = ½I(v/r)² + ½mv²
Now, putting the values in the equation, we have
5 * 9.8 * 0.3 = ½ * 10 * (v/2)² + ½ * 5 * v²
14.7 = 1.25 v² + 2.5 v²
14.7 = 3.75 v²
v² = 14.7/3.75
v² = 3.92
v = √3.92
v = 1.98 m/s
Thus, the speed is 1.98 m/s
The original Coulomb force between the charges is:
Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.
The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive.
Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
Q₂ is decreased to Q₂₂=(1/2)*Q₂=Q₂/2.
New force:
Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge
Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:
Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,
Fc₁=(k*(1/6)*Q₁*Q₂)/r²
Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:
Fc₁=(1/6)*Fc
So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc.
