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2 years ago

3. The expression 0.62 x10^3 is equivalent to...

Physics

2 answers:

Korolek [52]2 years ago

5 0

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

Nitella [24]2 years ago

5 0

Answer:

620

Explanation:

0.62*1000=620

10^3= 10*10*10=1,000

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Answer:

490.5 N

Explanation:

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F=kmg=0.5*100*9.81=490.5 N

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3 years ago

Read 2 more answers

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig

atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = $\sqrt{x^{2} + y^{2} }$

Then, the module of c will be:

module of c = $\sqrt{(xa + xb)^{2} + (ya + yb)^{2}}$ = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = $\sqrt{xa^{2} + ya^{2} }$ = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = $\sqrt{xa^{2} + ya^{2} }$ = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = $\sqrt{xb^{2} + yb^{2} }$ = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = $\sqrt{(2342 mi)^{2} + (234.4 mi)^{2} }$ = 2353.7 mi

The plane is 2353.7 mi from the starting position.

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Answer:

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Explanation:

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2 years ago

5. Forces have

Verdich [7]

In physics, forces are interactions that are able to change the velocity of an object.

Force is a vector quantity, so it has a magnitude and a direction.

The SI units of the force is the Newton (N).

Whenever an unbalanced force is applied to an object, the object experiences an acceleration, according to Newton's second law of motion:

$F=ma$

where

F is the force

m is the mass of the object

a is its acceleration

So, the acceleration of an object is proportional to the force applied:

$a=\frac{F}{m}$

In physics, arrows are used to represent vector quantities. Therefore, they are also used to represent forces.

In particular, when a vector quantity is represented by an arrowr:

- The length of the arrow is proportional to the magnitude of the vector quantity

- The direction of the arrow corresponds to the direction of the vector quantity

Therefore, if a force is represented through an arrow:

- The length of the arrow shows the strength (magnitude) of the force

- The direction of the arrow shows the direction of the force

As we said in part 5), the SI units of the force is the Newton (N).

We can rewrite the Newton in terms of fundamental units only. We can do it starting from the equation:

$F=ma$

where

F is the force

m is the mass

a is the acceleration

- The mass is measured in kilograms (kg)

- The acceleration is measured in meters per second squared ( $m/s^2$ )

Therefore, 1 N corresponds to:

$[N]=[kg][\frac{m}{s^2}]=[kg\cdot m \cdot s^{-2}]$

Gravity is an attractive force that exists between all objects that have mass. See more explanations about gravity in part 4).

Mass is a scalar quantity; it gives us a measure of the "amount of matter" contained in an object.

The SI unit of the mass is the kilogram (kg).

Being a scalar, mass has no direction, but only a magnitude.

Moreover, the mass is an intrinsec property of an object: therefore, it does not depend on the location of the object. So, an object has always the same mass, either it is on Earth or on another planet.

On the other hand, the force of gravity on an object depends on its location, so it changes.

As we said in part 3), gravity is an attractive force that exists between all objects that have mass.

The magnitude of the force of gravity between two objects is given by the Universal Law of gravitation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

From the equation above, we observe that:

- all objects are attracted to one another with a gravitational force that is proportional to the mass of the objects and inversely proportional to the square of the distance between them.

And so:

a. When the mass of one or both objects increases, the gravitational force between the objects increases

b. When the distance between two objects increases, the attraction between the objects decreases

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3 years ago