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Gennadij [26K]
3 years ago
15

A stone is thrown straight up from the ground with an initial speed of 44 m/s.At the same instant, a stone is dropped from a hei

ght of hmeters above ground level. The two stones strike the ground simultaneously. Find the heighth. The acceleration of gravity is 9.8 m/s2.Answer in units of m.
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer:

394.26 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v=u+at\\\Rightarrow 0=44-9.8\times t\\\Rightarrow \frac{-44}{-9.8}=t\\\Rightarrow t=4.49 \s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=44\times 4.49+\frac{1}{2}\times -9.8\times 4.49^2\\\Rightarrow s=98.77\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 98.77=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{98.77\times 2}{9.8}}\\\Rightarrow t=4.49\ s

Total time taken by the stone to reach the ground is 4.49+4.49 = 8.97 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{h\times 2}{9.8}}

The times are equal so,

8.97=\sqrt{\frac{h\times 2}{9.8}}\\\Rightarrow h=\frac{8.97^2\times 9.8}{2}\\\Rightarrow h=394.26\ m

The height is 394.26 m

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Answer:

The population mean are the same

Explanation:

Answer:

The population means are the same.

Explanation:

The hypotheses for a difference in two population means are similar to those for a difference difference two population proportions.

At null point, Ha=0

Let the mean population of the young one be u1

Let the mean population of the old one be u2.

Then, the difference between their mean population distance is given as

Ha=u2-u1

Since, Ha is null point, Ha=0

0=u2-u1

u2=u1

This shows that the mean population distance of the old is equal to the mean population distance of the young.

Therefore their mean population distance is the same

Since it is null alternative then, the population mean are the same.

We must sample the population using

1. Samples must be random to remove or minimize bias.

2. Sample must be representative of the populations in question.

8 0
3 years ago
0.0884 moles of a diatomic gas
Sloan [31]

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

\Delta\ U = nC_{v}\Delta\ T

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J

Now, we can apply First Law of Thermodynamics as follows:

\Delta\ Q = \Delta\ U + W

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\

<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>

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3 years ago
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A sample of gold has a mass of 30.94 grams and density of 19.32g/cm^3. What volume of space will this sample of gold occupy?
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Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3

Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3

Then, the answer is the option C. 
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2 years ago
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The sum of kinetic energy of molecules in a body is ______​
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Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
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