Answer:
Correct answer: Wave's frequency f = 7.5 · 10¹⁴ Hz
Explanation:
Given: wavelength λ = 4.0 · 10⁻⁷ m and
ultraviolet wave's speed V = 3 · 10⁸ m/s
The formula for calculating frequency is:
f = V / λ = 3 · 10⁸ / 4.0 · 10⁻⁷ = 0.75 · 10¹⁵ = 7.5 · 10¹⁴ Hz
f = 7.5 · 10¹⁴ Hz
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Relative dating is used to arrange geological events….
Relative dating puts geologic events in chronological order without requiring that a specific numerical age be assigned to each event….
Relative Dating uses the half life of isotopes to get the exact age of a rock or mineral.
Then the speed of the rider is, v = 2m/s
The linear velocity of an object in a straight line is its speed.
The following is the relationship between linear and angular velocity:
v = rω
The circumference of the bicycle wheel is,
2r = 2
r = 0.318m
The angular velocity of the wheel is,
ω = 1rev/sec = 2rad/s
Then the speed of the rider is,
v = rω
v = 0.318×2π
v = 2m/s
Learn more about angular velocity here:
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Answer:
a) 4.0 rad/s2
Explanation:
- For rigid bodies, Newton's 2nd law becomes :
τ = I * α (1)
where τ is the net external torque applied, I is the rotational inertia
of the body with respect to the axis of rotation, and α is the angular
acceleration caused by the torque.
- At the same time, we can apply the definition of torque to the left side of (1), as follows:
where τ = external net torque applied by Fnet, r is the distance
between the axis of rotation and the line of Fnet, and θ is the
angle between both vectors.
In this particular case, as Fnet is applied tangentially to the disk, Fnet
and r are perpendicular each other.
- Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:
Answer:
(a). 19.0 m/s
(b). 9.5m/s
Explanation:
Assuming speed of sound is 343m/s.
(a).
As the train approaches, from the the Doppler equation we have
solving for we get:
.
And as the train cuts its speed in half, the equation gives
substituting the value of we get:
or 68.6 km per hour, which is the speed of the train before slowing down.
(b).
The speed of the train after slowing down is half the previous speed; therefore,
or 34.3 km per hour.