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Anika [276]
2 years ago
5

What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a

nswer to two significant figures and include the appropriate units.
Physics
1 answer:
Umnica [9.8K]2 years ago
5 0

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec              

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity g=9.8m/sec^2

From first equation of motion we know that v=u+gt

v=0+9.8\times 9=88.2m/sec

So instantaneous velocity after 9 sec will be 88.2 m/sec  

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What voltage is required to move 6A through 20?<br><br>​
Marina CMI [18]

Answer:

120V

Explanation:

Given parameters:

Current  = 6A

Resistance  = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

           V = IR

Where V is the voltage

            I is the current

            R is the resistance

Now, insert the parameters and solve;

    V  = 6 x 20  = 120V

7 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
Electric current in a solid metal conductor is caused by the movement of what ?
Vlad1618 [11]
Electric current in a solid metal conductor is caused by the movement of electric charge. An electric current is the flow of electric charge. An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. The flow of ions also contributes to an electric current in, for example, the electrolyte of an electrochemical cell. 
5 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the
bazaltina [42]
Look on this website http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html
4 0
2 years ago
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