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2 years ago

6

Water bending around a rock in the middle of a lake disrupts the water behind the rock. This is an example of which wave propert

y?
diffraction
reflection
refraction
interference

1 answer:

Vladimir [108]2 years ago

5 0

Answer:

I would guess C. Refraction, since thats where waves bend.

Explanation:

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A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots

Alenkasestr [34]

Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved

$L_i = L_f$

$mv \frac{L}{2} = (I_1 + I_2)\omega$

here we will have

L = 0.250 m

v = 385 m/s

m = 1.90 gram

now moment of inertia of the plate will be

$I_1 = \frac{ML^2}{3}$

$I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2$

$I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2$

now from above equation

$0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega$

$\omega = 5.85 rad/s$

8 0

2 years ago

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

$s=ut+\frac{1}{2}*gt^{2} \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m$

Arrow misses the target by 5.843m ig the arrow us split horizontally

4 0

2 years ago

A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across

Rina8888 [55]

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

8 0

2 years ago

A ship sends out a 1200 Hz sound wave, which has a wavelength of 120 cm in the water. What would happen if that ship sent out a

OlgaM077 [116]

Answer:

the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

Explanation:

From wave,

v = λf................ Equation 1

Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: f = 1200 Hz, λ = 120 cm = 1.2 m

Substitute into equation 1

v = 1200(1.2)

v = 1440 m/s.

When the ship sent out a 600 Hz sound wave,

make λ the subject of formula in equation 1

λ = v/f............. Equation 2

Given: f = 600 Hz, v = 1440 m/s

Substitute into equation 2

λ = 1440/600

λ = 2.4 m or 240 cm.

When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

3 0

2 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago