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Fofino [41]
2 years ago
5

1. Which wave phenomenon is illustrated by this image?

Physics
2 answers:
Arisa [49]2 years ago
7 0

Answer:

Explanation: O B. Diffraction

balu736 [363]2 years ago
3 0
B.) diffraction
hope i’m right
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What is an object’s acceleration if it is moving at 30 m/s and comes to a stop in 5 s? –30 m/s2 –6 m/s2 30 m/s2 6 m/s2
Tanya [424]

Answer:32m/s

Explanation:

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3 years ago
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A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let 'F_A' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

F_A × 10.0 m - W × 4.0 m = 0

∴ F_A × 10.0 m = W × 4.0 m = 600 N × 4.0 m

F_A × 10.0 m = 600 N × 4.0 m

∴  F_A = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', F_A = 240 N.

8 0
3 years ago
A single sports fan is capable of yelling at an intensity level of 80 dB from a given distance. If 10,000 similar fans were all
riadik2000 [5.3K]

Answer: option "A" is correct

Explanation: the closet to understand the intensity level will be from 80db (decibels) upwards. So option a which is 84db is the answer. Thank you.

4 0
3 years ago
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How much longer will a 300 m steel bridge be on a 30C day than a –10C night?
RideAnS [48]

Answer:

Explanation: A bar of an unknown metal has a length of 0.975 m at 45°C and a length of 0.972 m at. 23°C.

6 0
3 years ago
A car is driving at 85 km/h and the driver spots a stop sign ahead. What coefficient of friction is needed to stop the car at th
almond37 [142]

Answer:

μ = 0.0315

Explanation:

Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)

N = m*g (1)

The friction force is equal to the product of the normal force by the coefficient of friction.

F = μ*N (2)

This way replacing 1 in 2, we have:

F = μ*m*g (2)

Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

\frac{1}{2} *m*v_{i}^{2}-(F*d)=  \frac{1}{2} *m*v_{f}^{2}

where:

vf = final velocity = 0

vi = initial velocity = 85 [km/h] = 23.61 [m/s]

d = displacement = 900 [m]

F = friction force [N]

The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.

Now replacing:

(1/2)*m*(23.61)^2 = μ*m*g*d

0.5*(23.61)^2 = μ*9,81*900

μ = 0.0315

5 0
3 years ago
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