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jenyasd209 [6]
3 years ago
6

Que fracción de luz que llega directa a una lamina de vidrio se refleja en la primera superficie?

Physics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

unos 40 o30 pienso

Explanation:

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You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m
sergey [27]

Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

  • mass of vehicle, m=1350\ kg
  • radius of curvature, r=71\ m
  • coefficient of friction, \mu=0.23

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

where:

\mu= coefficient of friction

N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

v=12.65\ m.s^{-1} is the maximum velocity at which the vehicle can turn without skidding.

5 0
3 years ago
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
What is the force due to gravity of a 38 kg student?
alukav5142 [94]

Answer:

F_g = 372.78 N

Explanation:

Formula for force of gravity is;

F_g = mg

Where;

m is mass

g is acceleration due to gravity

We are given;

Mass = 38 kg

Acceleration due to gravity has a constant value of 9.81 m/s²

Thus;

F_g = 38 × 9.81

F_g = 372.78 N

6 0
3 years ago
Help me plz i'll mark brainliest
HACTEHA [7]

Answer:

a PDF is what u use to upload an assignment to turn it in to get graded

8 0
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