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3 years ago

9

A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an

expression for the amount of time it takes the ball to travel from h to h/2.

1 answer:

agasfer [191]3 years ago

8 0

Answer:

$t=\sqrt{h/g}$

Explanation:

We use the kinematics equation to solve this question:

$y(t)=y_{o}+v_{o}t+1/2*a*t^{2}$

$v_{o}=0$ because the ball is dropped

$a=-g$ the acceleration is the gravity, negative because it points downwards

$y_{o}=h$ initial height

$y(t)=h/2$ final height

So:

$h/2=h-1/2*g*t^{2}$

$t=\sqrt{h/g}$

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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i

NikAS [45]

Answer:

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

Explanation:

As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

$v_f^2 - v_i^2 = 2 a d$

since we released it from rest so we have

$v_i = 0$

so here we have

$v_f = \sqrt{2gd}$

now if the distance is double then we have

$v_f' = \sqrt{2g(2d)}$

now from above two equations we can say that

$v_f' = \sqrt2 v_f$

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

4 0

3 years ago

Ohm’s Law is represented by the equation I=V/R. Explain how the current would change if the amount of resistance decreased and t

vesna_86 [32]

Answer:

The current will increase with reduction in the resistance.

Explanation:

Electrical resistance reduces the flow of electricity through a conductor just like friction reduces our speed. The higher the resistance the harder it will be for the current to flow and vice versa, hence, higher resistance produces a smaller current if the voltage is held constant. The voltage is the electrical drive.

3 0

3 years ago

Read 2 more answers

A laser emits a beam of light whose photons all have the same frequency. When the beam strikes the surface of a metal, photoelec

seraphim [82]

Answer:

the no. of ejected electrons per second will increase.

Explanation:

In photoelectric effect, when a light is incident on a metal surface it ejects some electrons from the metal surface. The energy of photon of light must be equal to or greater than the work function of that metal. All the extra energy above the work potential appears as the kinetic energy of the ejected electrons. So, greater he energy of photon greater will be the kinetic energy of the ejected electrons.

A single photon interacts with a single electron and ejects it only if its energy is greater than work function. So, the increase in no. of photons per second means an increase in the intensity of laser beam. And greater no. of photons, will interact with greater no. of electrons. So, <u>the no. of ejected electrons per second will increase.</u>

3 0

3 years ago

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as , where a

Misha Larkins [42]

Answer:

3 times louder

Explanation:

The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².

Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹

and I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₁ = 10㏒(I₁/I₀)

L₁ = 10㏒(10⁻⁹/10⁻¹²)

L₁ = 10㏒(10³)

L₁ = 3 × 10㏒10

L₁ = 30㏒10

L₁ = 30 dB

Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₂ = 10㏒(I₁/I₀)

L₂ = 10㏒(10⁻³/10⁻¹²)

L₂ = 10㏒(10⁹)

L₂ = 9 × 10㏒10

L₂ =90㏒10

L₂ = 90 dB

So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3

So, Braylee's music is 3 times louder than Jessica's music

6 0

3 years ago

A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

8 0

3 years ago