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3 years ago

5

Sedimentary rocks are formed due to the

1 answer:

Sonbull [250]3 years ago

3 0

Sedimentary rocks are formed due the following factors: weathering that breaks the rocks into smaller pieces or sediments, deposition of sediments, erosion (combination of weathering and movement of the resulting sediments), lithification (the changing of sediments into rock) and deposition of sediments.

According ths, sedimentary rocks are formed due to the:

B. weathering of rocks

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What is the formula of two vectors which are mutually perpendicular?

Svetllana [295]

Answer:

a .(b+c) = a.b+a.c

Explanation:

As the three vectors are perpendicular to each other, one of them must lie perpendicular to the plane containing the other two vectors…say a lies perpendicular to the plane of b and c.(b + c )gives another vector which lies in same plane of c and b .

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while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o

sveticcg [70]

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

$\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}$

The minimum radius not to exceed the centripetal acceleration is 2401 m.

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2 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, $q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C$

Charge 2, $q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C$

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}$

r = 0.8009 meters

or

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

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3 years ago

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

AVprozaik [17]

To solve this problem it is necessary to take into account the concepts related to the magnetic moment and the torque applied over magnetic moments.

For the case of the magnetic moment of a loop we have to,

$\mu = IA$

Where

I = Current

A = Area of the loop

Moreover the torque exerted by the magnetic field is defined as,

$\tau = IAB$

Where,

I = Current

A = Area of the loop

B = Magnetic Field

PART A) First we need to find the perimeter, then

$P = 2\pi r$

$r = \frac{P}{2\pi}$

$r = \frac{1.9}{2\pi}$

$r = 0.3025m,$

The total Area of the loop would be given as,

$A = \pi r^2$

$A = \pi 0.3025^2$

$A = 0.287m^2$

Substituting at the equation of magnetic moment we have

$\mu = (16*10^{-3})(0.287)$

$\mu = 4.58*10^{-3} A.m^2$

Therefore the magnetic moment of the loop is $4.58*10^{-3}Am^2$

PART B) Replacing our values at the equation of torque we have that

$\tau = IAB$

$\tau = (16*10^{-3})(0.287)(0.790)$

$\tau = 3.62*10^{-3}Nm$

Therefore the torque exerted by the magnetic field is $3.62*10^{-3}Nm$

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