Since it is dropped, initial velocity u = 0
Using s = ut + (1/2)gt^2, putting u = 0, g = 10 m/s^2
s =(1/2)gt^2, t = 3s
s = 0.5 * 10 * 3 *3
s = 45 m.
The bridge is 45m above the water.
Answer:
a.96cm
b.77cm
Explanation:
Highest intensity,
=19cm
Lowest intensity,
=29cm
48cm is the distance for half-wave, wavelength =48cm*2=96cm
b. From a, above we know that half-wavelength is 48cm
Also, the distance of known maximum is 29cm
Next maximum is calculated as:

The next maximum will occur at 77cm
Answer:
Given that
P = RT/V + a/V²
We know that
H= U + PV
For T= Constant (ΔU=0)
ΔH= ΔU +Δ( PV)
ΔH= Δ( PV)
P = RT/V + a/V²
P V= RT + a/V
dH/dV = d(RT + a/V)/dV
dH/dV = - a/V²
So the expression of dH/dV

b)
In isothermal process
(ΔU=0)
Now by putting the all values


ΔH = 17.06 L.atm
The force will be doubled if direction of one current is reversed and doubled.
The force between two parallel currents I₁ and I₂ that are spaced apart by r is measured as the ratio F/L. If the currents are flowing in the same direction, the force is attracting; if not, it is repulsive.
With a distance of 4.00 cm between them, two parallel wires repel one another with a force of 2.00 × 10⁻⁴ N/m. One wire is carrying 5.00 A of current.
I₁ = 5 A
d = 4 cm
F = 2 × 10⁻⁴ N/m
We have, F/L = μ₀ I₁ I₂ / L
Now, we know the force is directly proportional to the product of the current, and the direction of one current is reversed and doubled.
The force of interaction would become attractive and its magnitude would double as a result of switching the directions of either of the currents and doubling their size.
Learn more about force here:
brainly.com/question/12970081
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Answer:

Explanation:
To propose the solution of this problem we must use the relationship between work and energy
W = ΔK
the change in kinetic energy is
ΔK = K_f -K₀
ΔK = ½ m v_f² - ½ m v₀² = ½ m (v_f² - v₀²)
in this case
ΔK =
we can find work with the first law of thermodynamics
= Q + W
where \Delta E_{int} is the internal energy of the body, usually measured in the form of an increase in the temperature of the system
W = \Delta E_{int} - Q
if we consider that the internal energy does not change
W = -Q
we substitute everything in the first equation
-Q =
Because they are squared, the variables are positive, therefore, for the equation to be fulfilled, the exit velocity must be less than the entrance velocity.
