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3 years ago

(science) Just checking meh answer will mark brainliest

Physics

2 answers:

Katen [24]3 years ago

8 0

Answer:

D. Plate movement

Explanation:

Since they cause earthquakes...

Finger [1]3 years ago

3 0

Plate movement also known as tectonic plates

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a bus travels 4 km due north and 3 km due west going from bus station a to bus station b. the magnitude of the bus displacement

ladessa [460]

Answer:

5km

Explanation:

Magnitude of displacement is found by getting the resultant. Resultant is same as the bypotenuse hence

$R=\sqrt {{x^{2}+y^{2}}$ where x is the displacement in west direction and y is displacement in North direction. Substituting x with 3km and y with 4 km then

$R=\sqrt {{3^{2}+4^{2}}=5km$

4 0

3 years ago

Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but th

Shkiper50 [21]

The force required to extend a rod increases as the cross sectional area

increases.

The rod that experiences the largest force is <u>rod B</u>

Reason:

The elongation of a rod by the application of a force is given by the

following formula;

$\Delta L = \dfrac{F \cdot L}{A \cdot E}$

From the above equation, we have that the elongation is inversely

proportional to the cross sectional area, such that the extension of a rod by

a given force reduces as the cross sectional area of the rod increases.

Therefore, the force required to extend the length of a rod by a specific

amount increases as the cross sectional area of the rod increases,

indicating that the rod with the largest cross sectional area require the

most force and therefore, experiences the largest force.

The rod that experiences the largest force is the rod with the largest cross

sectional area, which is <u>rod B</u>

Learn more here:

brainly.com/question/12937199

4 0

2 years ago

A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje

Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

$\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m$

6 0

3 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

3 years ago

charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel

IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁ + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³ N / C .

3 0

3 years ago