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2 years ago

13

A ball rolls off an 8.0 m high building and strikes the ground 5.0 m away from the base of the building. How fast was the ball r

olling?

1 answer:

HACTEHA [7]2 years ago

4 0

Answer: horizontal speed is 3.9 m/s

Explanation: when ball starts to drop, its vertical speed v0 is zero.

We can calculate dropping time from s = v0t +0.5gt².

Dropping time t= √(2s/g)= √((2·8.0 m)/9.81 m/s²)= 1.277 s

Because ball travels horizontal distance s= 5.0 m

HorizontalSpeed v = s/t = 5.0 m/1.277s= 3,915 m/s

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In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

agasfer [191]

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$ .

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ = energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$ .

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2 years ago

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Answer:

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Explanation:

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2 years ago

The interference pattern seen when light passes through narrow, closely spaced slits, is due to

Lelu [443]

<h2>Answer: Diffraction</h2><h2 />

Diffraction is a characteristic phenomenon that occurs in all types of waves .

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2 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

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2 years ago

A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik

shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle $=40^{\circ}$

$x =12\pm 0.27$

y=1.05

equation of trajectory of ball is given by

$y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }$

for x=12.27

$1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }$

u=10.69

for x=11.73

$1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }$

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

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2 years ago