the relation that relates the speed of wave, frequency of wave and wavelength is given as
wavelength = 
the speed of wave in a medium remains constant. hence the wavelength is inversely related to the frequency of wave.
that means, as the frequency is increased, the wavelength decreases and vice versa.
hence the correct choice is
B decreases
Answer:
Explanation:
a )
momentum of baseball before collision
mass x velocity
= .145 x 30.5
= 4.4225 kg m /s
momentum of brick after collision
= 5.75 x 1.1
= 6.325 kg m/s
Applying conservation of momentum
4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.
v = - 13.12 m / s
b )
kinetic energy of baseball before collision = 1/2 mv²
= .5 x .145 x 30.5²
= 67.44 J
Total kinetic energy before collision = 67.44 J
c )
kinetic energy of baseball after collision = 1/2 x .145 x 13.12²
= 12.48 J .
kinetic energy of brick after collision
= .5 x 5.75 x 1.1²
= 3.48 J
Total kinetic energy after collision
= 15.96 J
Answer:
3) Ep = 13243.5[J]
4) v = 17.15 [m/s]
Explanation:
3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.
m = mass = 90 [kg]
h = elevation = 15 [m]
Potential energy is defined as the product of mass by gravity by height.
![E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D90%2A9.81%2A15%5C%5CE_%7Bp%7D%3D13243.5%5BJ%5D)
This energy will be transformed into kinetic energy.
Ek = 13243.5 [J]
4) The velocity can be determined by defining the kinetic energy, as shown below.
![E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2AE_%7Bk%7D%20%7D%7Bm%7D%20%7D%5C%5C%20v%3D%20%5Csqrt%7B%5Cfrac%7B2%2A13243.5%20%7D%7B90%7D%20%7D%5C%5Cv%3D17.15%5Bm%2Fs%5D)
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,
Ep=m*g*h=0.75*9.8*1.5=11.025 J
Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,
E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J
Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek
Ek=(1/2)*m*v²=12.5 J.
Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,
Ep=m*g*h=0.7*9.8*7=48.02 J
The order of examples starting with the lowest energy:
1. book, 2. ball, 3. stone, 4. brick