41.2 = h-1/2g(t-1)^2
<span> {-h = -1/2gt^2-1/2g+g*t-41.2
</span><span> {h = 1/2gt^2
</span><span> summing them up
</span><span> 0 = -1/2g+g*t-41.2
</span><span> 41.2 +4.9 = g*t
</span><span> t = 46.1/9.8 = 4.70 sec
</span><span> h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m </span>
<span>-Polar.
-Temperate
-Arid
-Tropical
-Mediterranean
<span>-Tundra
Hope this helps
</span></span>
A)
R1 = 30/(7*60)
We are multiplying 7 with 60 because there are 60 seconds in 1 minute.
R1 = 30/420 = 0.0714 gallons/second
b) For this we need to express gallons to cubic meters.
1 gallon = 0.003785 m^3
R2 = 0.0714*0.003785 = 0.00027 m^3/s
c) V = R2*t where V is volume of some tank.
which means that
t=V/R2
t = 3698.8s or
t = 1.0274 hours
Answer:
0.21486 mm
Explanation:
The formula for the maximum intensity is given by;
I = I_o•cos²(Φ/2)
Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)
Where;
y is the distance from the central maximum
d is the distance between the slits
λ is the wavelength
L is the distance to the screen
Thus;
I = I_o•πdy/(λL)
We are given;
d = 0.05 mm = 0.5 × 10^(-3) m
λ = 540 nm = 540 × 10^(-9) m
L = 1.25 m
I/I_o = 50% = 0.5
From earlier, we saw that;
I = I_o•πdy/(λL)
We have I/I_o = 0.5
Thus;
I/I_o = πdy/(λL)
Plugging in the relevant values;
0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)
Making y the subject, we have;
y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))
y = 0.00021486 m
Converting to mm, we have;
y = 0.21486 mm
Answer:
9.9 s
Explanation:
mass (m) = 1000 kg
initial speed (u) = 90 km/h
final speed (v) = 45 km/h
relationship between the speed (v) of the boat and the frictional force (fk) ⇒ fk = 70v
- the acceleration of the system will be given by (a) =
- acceleration is also the first differential of velocity with respect to time,
a =
- therefore acceleration (a) = =
- recall that fk = 70v
(a) = = =
(a) = =
- integrating both side of the equatin we have
t =
- the time required for the boat to slow down = t = = - 9.9 s = 9.9 s