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3 years ago

9

Vector A = 50 m, 20°. Vector -3A would be equal to

1 answer:

Kruka [31]3 years ago

3 0

Answer:

Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.

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At a location, rainwater runs off the face of a cliff that is 15 meters high. How is the face of the cliff likely to change over

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Weathering by water would cut its top deeper than the middle

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A small boat is moving at a velocity of 3.35m/s when it is accelerated by a river current perpendicular to the initial direction

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Answer:

v =7.1 m/s

Explanation:

Given that

u = 3.35 m/s

t= 5 s

a= 0.75 m/s²

The final velocity = v

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

v = 3.35 + 0.75 x 5 m/s

v =7.1 m/s

Therefore the final velocity will be 7.1 m/s

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Read 2 more answers

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

So

$m = 25 * 12.5$

$m \approx 310$

6 0

3 years ago

A river 800m wide flows at the rate of 5km/h . A swimmer who can swim at 10km/h in still water wants to cross the river straight

LuckyWell [14K]

Answer:

At an angle of $30^{\circ}$

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

$R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}$

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3 years ago