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3 years ago

Numerical Problems:

Physics

1 answer:

dangina [55]3 years ago

6 0

Displacement=1200m
Time=4min=4(60)=240s

$\boxed{\sf Velocity=\dfrac{Displacement}{Time}}$

$\\ \sf\longmapsto Velocity=\dfrac{1200}{240}$

$\\ \sf\longmapsto Velocity=5m/s$

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True or false:slow movements of mantle rock called radiation transfer heat in the mantle.

RUDIKE [14]

The best answer between the two options would be the second choice B) FALSE.

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3 years ago

Please help me with a speech.The topic of my speech is Passion.

Vikentia [17]

Answer:

Explanation:

Passion

For me, standing on the summit of Mt Everest was the result of following a process. The process of mountaineering. I love mountaineering. I am passionate about it. I love the months of planning for an expedition, the months of sweating and training to prepare my body physically. The meticulous preparation of my equipment. Most of all I love the huge mental challenge I have to overcome before each climb to confront my own fear. All these reasons are why I climb, they are why I climbed Mt Everest and that is why I continue to climb.

Passion is an enormously powerful force. It gives us the strength to get through hard times and setbacks. It gives us strength to overcome our fears, to ignore what other people think of us, to be disciplined and make sacrifices in pursuit of our dreams. Passionate people do not want to take shortcuts – they consider that ‘learning the process’ is an important part of the journey.

In mountaineering it’s easy to spot those who are not passionate about the process. They want to stand on top of the mountain but they are not really interested in the process of climbing the mountain. I feel for these people. Success without hard work is a hollow, empty feeling. They never last long in the sport.

Just as in life, successful mountaineers are the ones who are passionate. They are not there just to stand on the summit. Their passion gives them the energy to work the hardest, fight the longest, and in the words of Winston Churchill “never, never. never give-up”.

3 0

3 years ago

What is the mass of 3 m3 of a substance having density 1200 kg/m3

adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

4 0

3 years ago

A 1050 W carbon-dioxide laser emits light with a wavelength of 10μm into a 3.0-mm-diameter laser beam. What force does the laser

avanturin [10]

The force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

The given parameters;

<em>power of the laser light, P = 1050 W</em>
<em>wavelength of the emitted light, λ = 10 μm </em>

The speed of the emitted laser light is given as;

v = 3 x 10⁸ m/s

The force exerted by the laser beam on a completely absorbing target is calculated as follows;

P = Fv

$F = \frac{P}{v} \\\\F = \frac{1050}{3\times 10^8} \\\\F = 3.5 \times 10^{-6} \ N$

Thus, the force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

Learn more here:brainly.com/question/17328266

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2 years ago

In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point

Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ X volume of the oil drop

volume of the oil drop (spherical) = (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

= (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0

3 years ago