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3 years ago

A standing wave is the result of: refraction diffraction reflection resonance

Physics

2 answers:

natita [175]3 years ago

3 0

A standing wave is the result of a reflection.

dusya [7]3 years ago

3 0

The answer is relection

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Which object would have the greatest acceleration?

Andre45 [30]

Answer:

Explanation:

A and C are balanced, B has a resultant force of 5N right, and D has a resultant force of 20N right.

4 0

3 years ago

Si un cuerpo adquiere una carga de -0,02 C, ha ganado o ha perdido electrones? Cuantos?

sammy [17]

Question :

If a body acquires a charge of -0.02 C, has it gained or lost electrons? Many?

Solution :

We know, charge gained is shown by negative sign.

Since, charged acquired is given as -0.02 C .

Therefore, it is body has gained electrons.

Now, number of electrons is given by :

$n = \dfrac{net\ charge}{charge \ on \ one \ electron}\\\\n = \dfrac{-0.02}{-1.60 \times 10^{-19}}\\\\n = 1.25\times 10^{17}$

Hence, this is the required solution.

7 0

3 years ago

A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a

sattari [20]

Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

11^2 + 11^2 = c^2

121 + 121 = c^2

242 = c^2

c = 15.56

Rounding the answer makes it 16 km for the hiker's magnitude of displacement.

5 0

3 years ago

A bike accelerates uniformly(from rest to a speed of 7 m/s over a distance of 40 m. Determine

marshall27 [118]

Answer:

$0.61 m/s^2$

Explanation:

The bike's acceleration can be found by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the bike

u is the initial velocity

a is the acceleration

s is the distance covered

For the bike in the problem,

u = 0

v = 7 m/s

d = 40 m

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{7^2-0}{2(40)}=0.61 m/s^2$

3 0

4 years ago

A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

vo=5.87m/s

7 0

3 years ago