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Nastasia [14]
3 years ago
9

The average velocity of a tennis ball is measured during serving and found to be 216 km/hr.

Physics
1 answer:
icang [17]3 years ago
7 0

Answer:

The time between hitting the ball and meeting the ground is t=0.33\: s

Explanation:

We have a semi parabolic motion here. We know that the initial speed 216 km/h or 60 m/s in the x-direction.

So we can use the following equation:

v=\frac{\Delta x}{\Delta t}

60=\frac{20}{t}

t=\frac{20}{60}

t=0.33\: s

hope it helps you!

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Answer:

1 B. Convert v from km/min to m/s ( show work and units

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3 years ago
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J
Lyrx [107]

Explanation:

Given that,

Wavelength of the light, \lambda=4170\ A=4170\times 10^{-10}\ m

Work function of sodium, W_o=4.41\times 10^{-19}\ J

The kinetic energy of the ejected electron in terms of work function is given by :

KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J

The formula of kinetic energy is given by :

KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s

Hence, this is the required solution.

7 0
3 years ago
an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le
storchak [24]
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

6 0
4 years ago
When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When
Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, \dfrac{dI}{dt}=0.0240\ A/s

Induced emf, \epsilon=12.4\ mV=12.4\times 10^{-3}\ V

Current, I = 1.5 A

Magnetic flux, \phi=0.00338\ Wb

The induced emf through the solenoid is given by :

\epsilon=L\dfrac{dI}{dt}

or

L=\dfrac{\epsilon}{(di/dt)}........(1)

The self inductance of the solenoid is given by :

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From equation (1) and (2) we get :

\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}

N is the number of turns in the solenoid

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N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}

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or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0
3 years ago
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zaharov [31]

Answer:

18.4615385 amps

Explanation:

The voltage V in volts (V) is equal to the current I in amps (A) times the resistance R in ohms (Ω):

5 0
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