Answer:
The correct answer is option D i.e. A and C
Explanation:
The correct answer is option D i.e. A and C
for proficient catching player must
- learn to absorbed the ball force
- moves the hang according to ball direction to hold the ball
- to catch ball at high height move the finger at higher position
- to catch ball at low height move the finger at lower position
Answer:
50 N.
Explanation:
On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:
![\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20F_N%20%3D%20F_g%20%26%20%3D%20ma%20%5C%5C%20%26%20%3D%20mg%5Cend%7Baligned%7D)
The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:
![\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_N%20%3D%20F_g%20%3D%20%285%5Ctext%7B%20kg%7D%29%2810%20%5Ctext%7B%20N%2Fkg%7D%29%20%3D%2050%20%5Ctext%7B%20N%7D)
In conclusion, the normal force acting on the block is 50 N.
Answer:
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Explanation:
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
![f' = \frac{v + v_o}{v+ v_s} f](https://tex.z-dn.net/?f=f%27%20%3D%20%5Cfrac%7Bv%20%2B%20v_o%7D%7Bv%2B%20v_s%7D%20f)
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
![E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D%20%5C%5C%5C%5CE%20%3D%20%5Cfrac%7B6.626%2A10%5E%7B-34%7D%20%2A3%2A10%5E%7B8%7D%7D%7B130%2A10%5E%7B-9%7D%7D%20%5C%5C%5C%5CE%20%3D%2015.291%2A10%5E%7B-19%7D%20%5C%20J)
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
![v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2K.E%7D%7Bm%7D%20%7D%20%5C%5C%5C%5Cv%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%2A8.563%2A10%5E%7B-19%7D%7D%7B9.11%2A10%5E%7B-31%7D%7D%7D%5C%5C%5C%5Cv%20%3D%201.371%20%2A10%5E%7B6%7D%20%5C%20m%2Fs)
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.