Answer:
a) f=0.1 Hz ; b) T=10s
c)λ= 36m
d)v=3.6m/s
e)amplitude, cannot be determined
Explanation:
Complete question is:
Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.
Given:
number of wave crests 'n'= 5
pass in a time't' 54.0s
distance between two successive crests 'd'= 36m
a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have
f=n/t
f= 5/ 54 => 0.1Hz
b)The time period of wave 'T' is the reciprocal of the frequency
therefore,
T=1/f
T=1/0.1
T=10 sec.
c)wavelength'λ' is the distance between two successive crests i.e 36m
Therefore, λ= 36m
d) speed of the wave 'v' can be determined by the product of frequency and wavelength
v= fλ => 0.1 x 36
v=3.6m/s
e) For amplitude, no data is given in this question. So, it cannot be determined.
Time=50s
speed=25m/s
Distance = speed×time
=25×50
=1250m
DISTANCE TRAVELLED IS =1250m
Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
