Answer:
5.1m
Explanation:
Given the following parameters
Speed v = 10m/s
Time = 3.0s
Required
Displacement of the tennis ball
Using the equation of motion
v² =u²-2gh (g i s negative due to upward motion of the ball)
Substitute the given values into the expression above;
0² = 10²-2(9.8)h
0 = 100-19.6h
19.6h = 100
h = 100/19.6
h= 5.1m
Hence the displacement of the ball is 5.1m
Answer:
speed = 1.24 × 10⁸m/s
frequency = 4.74 × 10¹⁴Hz
wavelength = 262nm
Explanation:
the speed of the helium-neon light in zircon is given by,
v = c / n
c = 3 × 10⁸m/s is the speed of light in vacuum (and in air)
n = 2.419 is the refractive index of diamond
v = 3 × 10⁸ / 2.419
= 1.24 × 10⁸m/s
(b) Frequency
The wavelength of the light in air is:
λ₀ = 632.8 × 10⁻⁹
The frequency of the light does not depend on the medium, so it is equal in air and in diamond. Therefore, we can calculate the frequency by using the speed of light in air and the wavelength in air:
f₀ = c / λ₀
= 3 × 10⁸ / 632.8 × 10⁻⁹
= 4.74 × 10¹⁴Hz
and the frequency of the light indiamond is the same:
f¹ = f₀ = 4.74 × 10¹⁴Hz
(c) Wavelength
To calculate the wavelength of the light in daimond, we can use the relationship between speed of light in diamond and frequency:
λ¹ = v / f¹
= 1.24 × 10⁸ / 4.74 × 10¹⁴
= 2.62 × 10⁻⁷m
= 262nm
Answer:
Electric field, E = 936.19 N/C
Explanation:
It is given that,
Charge 1, 
Charge 2, 
Distance between them, d = 3 mm = 0.003 m
Torque, 
Angle between electric field and line connecting the charge, 
We need to find the torque exerted on the dipole. The torque experienced by the dipole in the electric field is given by :
p is the dipole moment, 
E = 936.19 N/C
So, the magnitude of electric field on the dipole is 936.19 N/C. Hence, this is the required solution.
Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
Answer:
Acceleration a=0.5 m/s²
Explanation:
Given data
Mass m=12,900 kg=1.29×10⁴kg
Thrust of engine F=28,000 N=2.8×10⁴N
gravitational acceleration g=1.67 m/s²
To find
Acceleration
Solution
As we know that
The net force can be given as
From Newtons second law of motion we know that
