Answer:
As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time
Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)
Explanation:
Answer:
Average speed = (2x+3y)/5 miles/hour
Explanation:
Average speed is the total distance travelled per unit time.
Average speed = total distance/total time taken
Given;
1. Travelled x miles per hour for 2hours.
2 travelled y miles per hour for 3 hours.
Distance covered is;
1. Distance = speed × time
d1 = x × 2 = 2x
2. d2 = y × 3 = 3y
Total distance travelled = d1+d2 = 2x + 3y
Total time taken = t1+t2 = 2+3 = 5hours
Average speed = (2x+3y)/5 miles/hour
Fusion occurs in the Sun's core, releasing energy that is transferred outward. Once in the radiative zone, gamma rays are transferred by radiation. They are converted to other types of photons, which move into the convective zone, where they are transferred by convection. Finally, energy is emitted from the photosphere.
The solution for this is:
Power = Energy transferred / Time taken
Energy Transferred in one second ( Power) = mgh/s
= (1.2x10^6)(9.8)(50) = 588000000 J/s
Power = 588000000 W
Or
Power is work done / time
Work done in one second = [ rate of fall of mass]
gh = 1.2* *9.81*50 x 10^6 J/s
= 5.886e+8 W
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.
Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²
We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m
Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
= </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
