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.
Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L) (1)
200 = k(60 - L) (2)
(2)/(1): 2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.
Speed will be = squreroot ( 2*g*L)
L is length of pendulum
Answer:
17.71N/m
Explanation:
The period of the spring is expressed according to the expression;
m is the mass of the object
k is the force constant
Given
m = 5.50kg
T = 3.50s
Substitute into the formula;
Hence the force constant of the spring is 17.71N/m
As we know that range of the projectile motion is given by
here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be
so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree
