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3 years ago

What is the number of waves that pass a fixed ping in a given amount of time?

1 answer:

7 0

The number of waves that pass a fixed point in a given amount of time is wave frequency. Wave frequency can be measured by counting the number of crests (high points) of waves that pass the fixed point in 1 second or some other time period. The higher the number is, the greater the frequency of the waves. :)

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How does nuclear fusion affect the life cycle of a star?.

Fittoniya [83]

Answer:

As the main sequence star glows, hydrogen in its core is converted into helium by nuclear fusion. When the hydrogen supply in the core begins to run out, and the star is no longer generating heat by nuclear fusion, the core becomes unstable and contracts

Explanation:

4 0

2 years ago

A toy rocket moving vertically upward passes by a 2.0-m-high window whose sill is 8.0 m above the ground. The rocket takes 0.15

sveta [45]

Answer:

$v_i = 18.86 m/s$

Explanation:

As we know that the speed of the rocket is v1 and v2 at the bottom and top of the window

then we will have

$d = (\frac{v_1 + v_2}{2}) t$

$2 = (\frac{v_1 + v_2}{2})(0.15)$

$26.67 m/s = v_1 + v_2$

also we know that

$v_2 - v_1 = (-9.81)(t)$

$v_2 - v_1 = (-9.81)(0.15) = -1.47$

now we have

$v_2 = 12.6$

also we have

$v_1 = 14.1 m/s$

now if the sill of the window is at height 8 m from the ground then we have

$v_1^2 - v_i^2 = 2 a h$

$(14.1^2) - v_i^2 = 2(-9.81)(8)$

$v_i = 18.86 m/s$

8 0

3 years ago

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center

Bess [88]

Answer:

10.93 rad/s

Explanation:

If we treat the student as a point mass, her moment of inertia at the rim is

$I_r = mr^2 = 67.8*3.7^2 = 928.182 kgm^2$

So the system moment of inertia when she's at the rim is:

$I_1 = I_d + I_r = 274 + 928.182 = 1202.182 kgm^2$

Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center

$I_2 = I_d + 67.8*0.456^2 = 274 + 14.1 = 288.1 kgm^2$

We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:

$I_1\omega_1 = I_2\omega_2$

$\omega_2 = \omega_1\frac{I_1}{I_2} = 2.62*\frac{1202.182}{288.1} = 10.93 rad/s$

8 0

3 years ago

What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 N at an angle of 35Â° above the

Olegator [25]

The acceleration of the wagon along the ground is 3.6 m/s².

To solve the problem above, we need to use the formula of acceleration as related to force and mass.

Acceleration: This can be defined as the rate of change of velocity.

⇒ Formula:

Fcos∅ = ma................. Equation 1

⇒ Where:

F = Force
∅ = angle above the horizontal
m = mass of the wagon
a = acceleration of the wagon

⇒ make a the subject of equation 1

a = Fcos∅/m..................... Equation 2

From the question,

⇒ Given:

F = 44 N
∅ = 35°
m = 10 kg

⇒ Substitute these values into equation 2

a = 44(cos35°)/10
a = 44(0.8191)/10
a = 3.6 m/s²

Hence, The acceleration of the wagon along the ground is 3.6 m/s²

Learn more about acceleration here: brainly.com/question/9408577

3 0

2 years ago

Electric circuits and electric current

lidiya [134]

an electric current is a flow of electric charge in electric circuits this is carried by moving electrons in a wire and an electric circuit is an electrical network of electrical components and model of interconnection consisting electrical elements

6 0

3 years ago