Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:
Given mass of box = 5.0 kg
Frictional force = 30 N
thus
Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero
Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.
Umm what are you trying to say
Answer:
835.29 Hz
Explanation:
When moving towards the source of sound, frequency will be given by
f*=f(vd+v)/v
Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.
Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then
f*=800(15+340)/340=835.29411764704 Hz
Rounded off, the frequency is approximately 835.29 Hz
Answer:
B. d(low)=4d(high)
Explanation:
Frequency of a string can be written as;
f = v/2L
Where;
v = sound velocity
L = string length
Frequency can be further expanded to;
f = v/2L = (1/2L)√(T/u) ......1
Where;
m= mass,
u = linear density of string,
T = tension
p = density of string material
A = cross sectional area of string
d = string diameter
u = m/L .......2
m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)
f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))
f = (1/2L)√(T/((p(πd^2)L/4)/L))
f = (1/2L)√(4T/pπd^2)
f = (1/L)(1/d)√(4T/pπ)
Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.
f ~ 1/d
So, if
4f(low) = f(high)
Then,
d(low) = 4d(high)
