Answer:
Explanation:
Total heat = Work done = Force × distance
distance = 0.075 × 12 = 0.9 m
W = 45 × 0.9 = 40.5 joules
Specific heat of the human hand = 3.5 kj/kg = 3.5 j/g
Q = MCΔT
ΔT = (Q) ÷ (MC)
ΔT = 40.5 ÷ (3.5 × 1) = 11.57°C
Answer:
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t], where the origin is at the left end of the string, the x-axis is along the string, and the y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation y(x, t) for this string if it were vibrating in its eighth harmonic?
The stone's altitude at time 
is given by
where 
is the acceleration due to gravity. The stone reaches the ground when 
:
If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.
Answer:
mph
Explanation:
= Speed of bird in still air
= Speed of wind = 44 mph
Consider the motion of the bird with the wind
= distance traveled with the wind = 9292 mi
= time taken to travel the distance with wind
Time taken to travel the distance with wind is given as
eq-1
Consider the motion of the bird with the wind
= distance traveled against the wind = 6060 mi
= time taken to travel the distance against wind
Time taken to travel the distance against wind is given as
eq-2
As per the question,
Time taken with the wind = Time taken against the wind
mph
