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1 year ago

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brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o

f the child as she swings?

1 answer:

beks73 [17]1 year ago

6 0

Maximum speed of the child as she swings is 2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>

The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation.
A periodic change in a substance's value between two values or around its central value is another way to define oscillation.

To learn more about Oscillation refer to:

brainly.com/question/28312746

#SPJ1

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The energy expenditure value of traveling by car is 3.6 mj/passenger-kilometer. The value for traveling by train is 1.1 mj/passe

andrezito [222]

Answer:

Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.

Explanation:

Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.

Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:

$\eta_{engine} \cdot U_{engine} = \frac{1}{2} \cdot m_{car} \cdot v^{2} + \mu_{r} \cdot m_{car} \cdot g \cdot \Delta s$

7 0

3 years ago

the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert

AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure $P$ is the force $F$ exerted by a gas, a liquid or a solid on a surface (or area) $A$ , its unit is Pascal $Pa$ which is equal to $N/m^{2}$ and its formula is:

$P=\frac{F}{A}$ (1)

In this case we have the surface of a sole of a tennis shoe:

$A=0.0290 m^{2}$ (2)

And the mass $m$ of the person who wears it:

$m=65 kg$

On the other hand, we know the weight is the force $F$ the Earth exerts on people and objects due gravity $g$ :

$F=m.g=(65 kg)(9.8m/^{2})$

$F=637N$ (3)

Substituting (2) and (3) in (1):

$P=\frac{637N}{0.0290 m^{2}}$ (4)

Finally:

$P=21965.517 Pa$ This is the pressure the shoe exert on the ground

5 0

3 years ago

A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of e

olasank [31]

Answer:

Explanation:

Given

Train travels towards south with a velocity if $v_t=25\ m/s$

Rain makes an angle of $\theta =66^{o}$ with vertical

If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity

suppose $v_r$ is the velocity of rain with respect to ground then

$v_r\sin\theta =v_t$

$v_r\times \sin (66)=25$

$v_r=27.36\ m/s$

Therefore velocity of rain drops is 27.36 m/s

8 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

2 years ago

Where is the switch located on this diagram?

Fed [463]

For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.

6 0

2 years ago