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3 years ago

I throw a ball off the edge of a 15.0m cliff. If I threw it horizontally at 8.0 m/s, how much time did it take to fall?

1 answer:

3 0

Just ignore the horizontal component

if you have a vertical displacement of 15m, 0ms^1 initial velocity, end velocity is ignored, we know the acceleration due to gravity as 9.81ms^2 so we can work out the time using SUVAT

S=15
U=0
V=?
A=9.81
T=?

S=UT + 0.5 AT^2

UT=0

therefore,

S=0.5AT^2

rearrange to:

T=SQR (2S/A)

T = 1.75 seconds

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Consider the incomplete equation below

kati45 [8]

Answer:

Nuclear fusion produces elements that are heavier than helium.

Explanation:

6 0

3 years ago

Read 2 more answers

In a single-slit diffraction experiment, the slit is 0.1mm wide and is illuminatedby monochromatic light with a wavelength of 50

dem82 [27]

Answer:

$0.1$ m

Explanation:

$d$ = slit width = 0.1 mm = 0.1 x 10⁻³ m

$\lambda$ = wavelength of monochromatic light = 500 nm = 500 x 10⁻⁹ m

$D$ = Distance of the screen = 10 m

$w$ = Spacing between successive minima

Spacing between successive minima is given as

$w = \frac{2D\lambda }{d}$

Inserting the values given

$w = \frac{2(10)(500\times10^{-9}) }{0.1\times10^{-3}}$

$w = 0.1$ m

4 0

3 years ago

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

$\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

We're told the water is leaking out at a rate of $10,500\,\frac{\mathrm{cm}^3}{\rm min}$ , so we find the rate at which it's being pumped in to be

$\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}$

$\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}$

4 0

3 years ago

In a house 5 bulbs of 25 watts each are used for 6 hours a day.Calculate the unit of electricity consumed in a month of 30 days.

Svetradugi [14.3K]

(5 bulbs) x (25 watt/bulb) x (6 hour/day) x (30 day/month) =

   (5 x 25 x 6 x 30) watt-hour/month =

   22,500 watt-hour/month .

The most common unit of electrical energy used for billing purposes
is the 'kilowatt-hour' = 1,000 watt-hours .

   22,500 watt-hour/month = <em>22.5 kWh/month</em>.

   (22.5 kWh/month) x (1.50 Rs/kWh) = <em>33.75 Rs / month

</em>

4 0

3 years ago

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula: In this equation stands for the Rydberg e

labwork [276]

Answer:

Wavelength, $\lambda=657\ nm$

Explanation:

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula as :

$E=\dfrac{-R}{n^2}$ .............(1)

Where

R is the Rydberg constant

n is the number of orbit

We need to find the wavelength of the line in the absorption line spectrum of hydrogen caused by the transition of the electron from an orbital with to an orbital with n₁ = 2 to an orbital with n₂ = 3.

Equation (1) can be re framed as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

$\dfrac{1}{\lambda}=1.096\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{3^2})$

$\lambda=6.569\times 10^{-7}\ m$

$\lambda=657\ nm$

So, the the wavelength of the line in the absorption line spectrum is 657 nm. Hence, this is the required solution.

3 0

3 years ago