Answer:
Maximum height the atmosphere pressure can support the
water=10.336 m
Explanation:
We know that ,
Case 1 - Mercury in the tube
Case 2 - Water in the tube
Since atmospheric pressure is same
.
or, 
∴ 
Hence height of the water column =10.336 m
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
Answer:
= 7.02 ° C
Explanation:
The liquid water gives heat to melt the ice (Q₁) maintaining the temperature of 0 ° C and then the two waters are equilibrated to a final temperature.
Let's start by calculating the heat needed to melt the ice
Q₁ = m L
Q₁ = 0.090 3.33 10⁵
Q₁ = 2997 10⁴ J
This is the heat needed to melt all the ice
Now let's calculate at what temperature the water reaches when it releases this heat
Q = M 
(T₀ -)
Q₁ = Q
= T₀ - Q₁ / M
= 20.0 - 2997 104 / (0.600 4186)
= 20.0 - 11.93
= 8.07 ° C
This is the temperature of the water when all the ice is melted
Now the two bodies of water exchange heat until they reach an equilibrium temperature
Temperatures are
Water of greater mass T₀₂ = 8.07ºC
Melted ice T₀₁ = 0ºC
M (T₀₂ - ) = m ( - T₀₁)
M T₀₂ + m T₀₁ = m + M
= (M T₀₂ + 0) / (m + M)
= M / (m + M) T₀₂
let's calculate
= 0.600 / (0.600 + 0.090) 8.07
= 7.02 ° C
Answer:
<em>A.The rabbits in the new habitat will have lower genetic variation than the rabbits in the original habitat. </em>
<em></em>
Explanation:
If two animals of opposite sex are isolated from a larger group of animal, and made to reproduce. They will produce offspring with similar genetic makeup. If this offspring still remain isolated, and continue to interbreed within themselves for a number of consecutive generations, their offspring will all be very closely related genetically. Situations like this just as with the two rabbits in the question leads to a lower genetic variation within the offspring of the two animals.
Animals need to reproduce within a larger group in order to increase genetic variation. Increasing genetic variation reduces the risk of been sucked into a gene pool. A lower genetic variation reduces the fitness of the animals involved. It is only an advantage in cases in which the the original pair are resistant to a deadly disease. In this case all the offspring also develop this immunity. Mostly the effects of a lower genetic variation leaves negative impacts, and animals try to avoid this by preferring to interbreed with unrelated partner
Ey = 375 cos [ kx - (92.20x10^14).t ]
<span>the formula of the electric field of an electromagnetic wave : </span>
<span>=> Ey = Emax cos (k x - ω t) </span>
<span>=> Ey = Emax cos [(2π/λ) - (2π f).t ] </span>
<span>then : (2π f).t = (92.20x10^14).t </span>
<span>===> 2π f = 92.20x10^14 </span>
<span>===> f = 92.20x10^14 / 2π </span>
<span>===> f = 1.46 x 10^15 hertz </span>
<span>the speed of electromagnetic wave : c = 3 x10^8 m/s </span>
<span>then : c = f λ </span>
<span>======> λ = c / f </span>
<span>======> λ = 3 x10^8 / 1.46 x 10^15 </span>
<span>======> λ = 2.0547 x 10^-7 meters = 0.205 μm</span>
