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Contact [7]
2 years ago
14

The sum of the interior angles of a quadrilateral is _______. 180° 90° 270° 360°

Physics
2 answers:
nlexa [21]2 years ago
7 0
Year9 formula: 180 (n-2)
Answer: 360 



olya-2409 [2.1K]2 years ago
4 0

Answer:

360°

Explanation:

Quadrilateral are polygons that have four sides. The sum of internal angle of a quadrilateral is 360°. The quadrilateral are closed shape with four sides. Examples of quadrilaterals are trapezium, rectangle, square, Rhombus and parallelograms.

Every quadrilaterals can be divided into two triangle. The sum of angle in a triangle is 180°. Adding the two triangle angle together , we will get the sum as 360°.

Sum of the interior angle of a polygon = 180(n-2) . In the case of quadrilateral, it has 4 sides. n represents the number of sides of the polygon .

interior angle of a quadrilateral = 180(4-2) = 360°

The image illustrate how the interior angle of quadrilaterals can be sum.

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Goryan [66]
The standard unit is KW/hr, = 1,000W/hr.
(85 + 60) = 145W.
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6 0
2 years ago
A block lies on a horizontal frictionless surface and
zhenek [66]

Answer:

0.1 m

Explanation:

F = Force exerted on spring = 3 N

k = Spring constant = 60 N/m

x = Displacement of the block

As the energy of the system is conserved we have

Fx=\dfrac{1}{2}kx^2

\\\Rightarrow x=\dfrac{2F}{k}

\\\Rightarrow x=\dfrac{2\times 3}{60}

\\\Rightarrow x=0.1\ m

The position of the block is 0.1 from the initial position.

6 0
3 years ago
Simple physics (Final) (Pic provided)
Igoryamba

Answer:

15m/s

Explanation:

Divide distance by time

3 0
3 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
Cocking your head would be most useful for detecting the ______ of a sound.
statuscvo [17]
Cocking your head would be most useful for detecting the LOCATION of a sound.
5 0
3 years ago
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