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I am Lyosha [343]
3 years ago
6

In a certain region the average horizontal component of Earth's magnetic field is 17 μT, and the average inclination or "dip" is

79°. What is the corresponding magnitude of Earth's magnetic field in microteslas?
Physics
1 answer:
Rufina [12.5K]3 years ago
6 0

Answer:

the Earth magnetic field in that region is about 89.09 \muT

Explanation:

If the magnetic field has an inclination of 79^o, and its horizontal component is 17 microTesla, then we can calculate the magnitude of the full magnetic field via the cosine function. Notice that the decomposition of a vector in horizontal and perpendicular components originates a right angle triangle where the vector is the hypotenuse of the triangle.

cos(\theta)=\frac{adjacent}{hypotenuse}\\cos(79^o)=\frac{17}{B} \\B=\frac{17}{cos(79^o)}\,\mu T  \\B=89.09\,\,\mu T

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Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
Four resistors (67 ohm, 83 ohm, 433 ohm, and 309 ohm in that order) are connected in series to a 7.92 V battery of negligible in
Whitepunk [10]

Answer:

.737 v

Explanation:

Since they are in series....they all have the same current running through them.....find the total resistance to calculate the current:

R = 67 + 83 + 433 + 309 = 892 ohm

V/R = current = 7.92 / 892 = 8.87 mAmps

Now the voltage across ecah resistor is   I R

 for the second one   8.87 ma * 83 ohm = V = .737 V

7 0
2 years ago
How much time does it take a person to walk 12km north at the velocity at a velocity of 6.5 km/hrs?
alukav5142 [94]

"6.5 km/hr" is not a velocity.  It's just a speed, so
we don't know what direction he's walking.

If he happens to be walking north, then it takes him

               (12 km) / (6.5 km/hr)  =  1.846... hours  (rounded) .

If he's walking in any other direction, it takes him longer than that.

If the angle between north and the direction he's walking is
90 degrees or more, then he can never cover any northward
distance, no matter how long he walks.

6 0
3 years ago
Read 2 more answers
(A) how much work is done when you push a crate horizontally with 100 N across a 10-m factory floor?
Mila [183]

Answer:

A) 1000 joules

Explanation:

In general work is given by the equation:

W=\intop_{a}^{b}\overrightarrow{F}\cdot d\overrightarrow{s} (1)

A) With \overrightarrow{s} the displacement and \overrightarrow{F} the force applied, because the force and the displacement are parallel (the crate is pushed horizontally) \overrightarrow{F}\cdot d\overrightarrow{s} is simply F\,ds, and because the path is a straight line and the force is constant work is:

W=FS (2),

W=(100)(10)=1000\,J

B) The work-energy theorem says that the total work on a body is equal to the change on kinetic energy:

W_{tot}=\varDelta K (3)

The total work on the crate is the work done by the push and plus the work of the friction W_{tot}=W + W_{f} (4) , as (A) because forces are parallel to the displacement W= FS (5) and W_{f}=-fS (6), the due friction always has negative sign because is opposite to the displacement, using (6), (5) and (4) on (3):

FS-fS=\varDelta K (3)

1000-(70)(10)=300\,J

C) The energy is lost by friction, so the amount of energy turned into heat is the work the friction does:

Q=fS=(70)(100)=700\,J (3)

6 0
3 years ago
A large semi-truck is moving a house from one lot to another. The amount of force required to move the house 15.A horizontally a
dlinn [17]

Answer:

252J

Explanation:

Given parameters:

Distance  = 72m

Force  = 3.5N

Unknown:

Work done on the house  = ?

Solution:

Work done is the force applied to move a body through a particular distance.

    Work done  = Force x distance

Now insert the parameters and solve;

   Work done  = 3.5 x 72  = 252J

5 0
2 years ago
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