Answer:
The height of Sears Tower is 1448.5 feet.
Explanation:
<h3>
We apply the free fall formula to the ball:
</h3><h3>
</h3><h3>y: The vertical distance the ball moves at time t </h3><h3>
i: Initial speed
</h3><h3>g=Gravity acceleration=
</h3>
Known information
We know that the vertical distance (y) that the ball moves in 9,5s is equal to height of Sears Tower (h).
Too we know that the ball is released from rest, then,
=0
Height of Sears Tower calculation:
We replace in the equation 1 the data following;
Answer: The height of Sears Tower is 1448.5 ft
Answer:
I=2A
R=5
Explanation:
formula
V=IR
=2x5
Voltage=10 volt
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First, find how many copper atoms make up the ball:
moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0.<span>77638</span><span>mol
</span> # of atoms = (0.77638 mol) (6.02 × 10^23 atoms per mol) = 4.6738*10^23<span> atoms </span>
<span> There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
</span> normal # electrons = (4.6738 × 10^23 atoms) (29 electrons per atom) = <span>
<span>1.3554</span></span><span>× 10^25 electrons
</span>
<span> Currently, the charge in the ball is 2.0 µC, which means -2.0 µC worth of electrons have been removed.
</span><span> # removed electrons = (-2.0 µC) / (1.602 × 10^-13 µC per electron) = 1.2484 × 10^13 electrons removed
</span><span> # removed electrons / normal # electrons = </span>
<span>(1.2484 × 10^13 electrons removed) / (1.3554 × 10^25 electrons) = 9.21 × 10^-13 </span>
<span> That's 1 / 9.21 × 10^13 </span>
Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.
