A child throws a stone vertically upwards with an initial velocity of 14.9 m/s. What is the maximum height of the stone?

A child pushes a toy box across the floor in 5 seconds. If he did 30 J of work on the toy box, what amount of power was required

s2008m [1.1K]

My answer is 6 watts because 30J/5s is 6

4 0

3 years ago

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I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT.

Tems11 [23]

Ultraviolet light with a wavelength shorter than visible light and a higher radiant energy than visible light.

The shorter the wavelength, the higher the frequency and energy.

6 0

3 years ago

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A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 hz. what driving freq

Korvikt [17]

600Hz is the driving frequency needed to create a standing wave with five equal segments.

To find the answer, we have to know about the fundamental frequency.

<h3>How to find the driving frequency?</h3>

The following expression can be used to relate the fundamental frequency to the driving frequency;

f(n) = n * f (1)

where, f(1) denotes the fundamental frequency and the driving frequency f(n).

The standing wave has four equal segments, hence with n=4 and f(n)=4, we may calculate the fundamental frequency.

f(4) = 4× f (1)

480 = 4× f(1)

f(1) = 480/4 =120Hz.

So, 120Hz is the fundamental frequency.

To determine the driving frequency necessary to create a standing wave with five equally spaced peaks?
For, n = 5,

f(n) = n 120Hz,

f(5) = 5×120Hz=600Hz.

Consequently, 600Hz is the driving frequency needed to create a standing wave with five equal segments.

Learn more about the fundamental frequency here:

brainly.com/question/2288944

#SPJ4

8 0

1 year ago

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

A 5kg ball is on top of the school building at a height of 40m above the ground.

mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0

2 years ago