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Vaselesa [24]
2 years ago
7

A boy threw a ball upward and it took 3 sec to reach the highest point. Determine the initial velocity of the ball.

Physics
1 answer:
makkiz [27]2 years ago
6 0

Answer:

V initial = 29.4 m.s²

Explanation:

( Using the laws of motion)

V final = V initial + Acceleration × time

0 = V initial + ( -9.8)(3)

29.4 = V initial

* I took upward as positive that's why I substituted -9.8 *

* for V final we know that at maximum height the ball is not moving thats why is = 0 *

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Galina-37 [17]
The current intensity is defined as the amount of charge Q that passes through a certain point of an electrical wire in a time interval of \Delta t:
I= \frac{Q}{\Delta t}

In our problem, the current intensity is
I=12.0 mA=0.012 A
while the amount of charge that passes a certain point of the wire is 
Q=6.50 C

If we re-arrange the previous equation:
\Delta t= \frac{Q}{I}
we can find the time needed for this amount of charge to pass through a point of the wire:
\Delta t= \frac{6.50 C}{0.012 A} =541.7 s
6 0
3 years ago
What is the value of the net electrostatic force on (a) particle 1 and (b) particle 2 due to the other particles
MrRissso [65]

Answer:

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Explanation:

7 0
2 years ago
an elevator suspended by a vertical cable is moving downward but slowing down. the tension in the cable must be:
vladimir1956 [14]

Answer:

F = M a is the vector equation involved

F = T - M g      are the forces acting on the elevator   (scalar equation)

T - M g = M a

T = M (a + g)    remember this a scalar

If a is slowing down then it must have a positive acceleration upwards

Therefor the tension in the cable must be greater than zero

When the tension increases to M g, a has increased to zero

For a to be zero, no acceleration, T = M g

6 0
2 years ago
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

4 0
3 years ago
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alina1380 [7]
The condition of earths atmosphere at a given time and place is the weather.
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