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fredd [130]
3 years ago
15

(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation​

Physics
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

the answee is

2.968456 ×10^7

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Consider two rocks with masses of 1 and 10 kilograms. What is the relation between their inertias? Between their masses? Between
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The greater mass the object has the greater its inertia would be. As inertia becomes greater, the same happens with the force which is needed stop motion. Linear motion and rotational motion are quite different, because the first one depends only on mass while the second embraces mass, size and shape of an object. According to the information I shared, one will not be able to stop 10 kg mass due to far greater inertia than object of 1kg mass.

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How do repeater stations improve the quality of a broadcast?
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They pick up broadcast signals, amplify them, then send them out.
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3 years ago
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
igor_vitrenko [27]

Answer:

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

b) Her highest height above the board is 0.82 m

c) Her velocity when her feet hit the water is 7.16 m/s

Explanation:

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=4-9.81\times t\\\Rightarrow \frac{-4}{-9.81}=t\\\Rightarrow t=0.41\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4\times 0.41+\frac{1}{2}\times -9.81\times 0.41^2\\\Rightarrow s=0.82\ m

b) Her highest height above the board is 0.82 m

Total height she would fall is 0.82+1.8 = 2.62 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.62=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.62\times 2}{9.81}}\\\Rightarrow t=0.73\ s

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

v=u+at\\\Rightarrow v=0+9.81\times 0.73\\\Rightarrow v=7.16\ m/s

c) Her velocity when her feet hit the water is 7.16 m/s

6 0
3 years ago
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12
Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

3 0
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A car travels 90 meters north in 15 seconds. Then turns around and travels at 40 meters south in 5.0 seconds. what is the averag
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Average velocity  =  (displacement) / (total time)

Displacement = distance and direction between
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The car travels 90 meters north, then 40 meters south. 
So it ends up 50 meters north of where it began. 

Displacement = 50 meters north
Total time  =  (15 + 5) = 20 seconds

Average velocity = (50 meters north) / (20 seconds)

                             =  2.5 m/s north
4 0
3 years ago
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