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worty [1.4K]
2 years ago
7

A water wave approaches a wall with a small hole in it. The wave is blocked by the wall, except for the part that goes through t

he hole. The waves coming out of the hole begin to spread out as the new source of wavefronts. This is an example of what property?
a Interference
b Amplitude
c Collision
d Diffraction
Physics
1 answer:
Eduardwww [97]2 years ago
7 0

Answer:

diffraction

Explanation:

took da test

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Two cars are traveling in the same direction and with the same speed along a straight highway. Does either driver hear a differe
Trava [24]

Answer:

No, either driver can not hear a different frequency from the other car's horn than they would if the cars were stationary.

Explanation:

Either driver hear a different frequency from the other car's horn than they would if the cars were stationary if two cars are traveling in the same direction and with the same speed along a straight highway because neither driver experiences a Doppler shift

8 0
3 years ago
ANSWER SOOONN PLZ!!
Ket [755]

c it is both a and b

3 0
3 years ago
You throw a 50.0g blob of clay directly at the wall with an initial velocity of -5.00 m/s i. The clay sticks to the wall, and th
Whitepunk [10]

Answer:0.25 kg-m/s

Explanation:

Given

mass of blob m=50 gm

initial velocity u=-5 m/s\ \hat{i}

time of collision t=20 ms

we know Impulse is equal to change in momentum

initial momentum P_i=mu

P_i=50\times 10^{-3}\times (-5)=-0.25 kg-m/s

Final momentum P_f=50\times 10^{-3}v

P_f=0 as final velocity is zero

Impulse J=P_f-P_i

J=0-(-0.25)

J=0.25 kg-m/s

5 0
3 years ago
4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
KatRina [158]

Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

8 0
3 years ago
A bus travels north on some busy city streets for 2.5 km, and a trip
d1i1m1o1n [39]

Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

Hence, the velocity of the  bus, V = 11.31 m/s

8 0
3 years ago
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