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andrey2020 [161]

3 years ago

8

pakka acquires a velocity of 72 km per hour in 10 seconds starting from rest find a ) the acceleration b)the average velocity c)

the distance travelled in this time

1 answer:

nadezda [96]3 years ago

8 0

Acceleration equals 24 km/s
Average equals 396km/s

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What is the speed of a wave with a wavelength of 3 m and a frequency of .1Hz?

Yuri [45]

We know,
Speed = Frequency * Wavelength
Speed = 3 * 0.1 m/s [hertz = 1/sec.]
So, your final answer is 0.3 m/s

Hope this helps!!

6 0

3 years ago

Read 2 more answers

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

if gas in a sealed container has a pressure of 50kpa at 300k what will the pressure be if the tempature rises to 360k

Zarrin [17]

The pressure law states that pressure is directly proportional to temperature.
p=kt where p is pressure, k is a constant, and t is temperature.

p=kt -- substitute
50000=k*300000
k=1/6
p=1/6*360000
p=60000 -- in pa not kpa

The pressure is 60kpa

3 0

3 years ago

Earth surface are composed of 70% of water and 30% of lands or soil.

musickatia [10]

Answer:

idk

Explanation:

7 0

3 years ago

Find the gravitational potential energy of an 84 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as

djverab [1.8K]

Answer:

$E=7.28\times 10^6\ J$

Explanation:

Given that,

Mass of a person, m = 84 kg

The person is standing at a top of Mt. Everest at an altitude of 8848 m

We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :

E = mgh, g is the acceleration due to gravity

$E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J$

So, the gravitational potential energy of the person is $7.28\times 10^6\ J$

6 0

3 years ago