two negative charge of -2.0 c and a positive charge of 3.0 c are separated by 80 m what is the force between two charges

A watermelon is thrown down from a skyscraper with a speed of 7.0\,\dfrac{\text m}{\text s}7.0 s m 7, point, 0, space, start f

Degger [83]

Answer:

y = 17,89 m

Explanation:

Let us fixate the reference point in top of the building, from where the watermelon is thrown down. We will assume also that the positive axis of our system points up. We describe the watermelon’s motion with the equation:

$v_y^2 =v_0^2 + 2ay$

Clearing the equation so we isolate y we have that:

$y = (v_y^2 - v_0^2 )/2a$

Making a substitution with the values from the statement we have:

$y = ((20 m/s)^2 - (7 m/s)^2)/(2*9,81 m/s^2) = 17,89 m$ ]

So, this skyscraper is about 17,89 m tall; which is not very tall for a skyscraper but who am I to judge. 17,89 m is also the displacement of the watermelon from the point it was thrown down.

I hope everything was clear with my explanation. If I can help with anything else, just let me know. Have an awesome day :D

7 0

2 years ago

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Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrat

lyudmila [28]

Answer:

$\theta_{min} = 1.21 \times 10^{-5}\ rad$

Explanation:

given,

diameter of the beam (d)= 5.85 cm

= 0.0585 m

average wavelength of the(λ) = 580 n m

angle of of spreading = ?

according to the Rayleigh Criterion the minimum angular spreading, for a circular aperture, is

$\theta_{min} = 1.22\ \dfrac{\lambda}{d}$

$\theta_{min} = 1.22\ \dfrac{580 \times 10^{-9}}{0.0585}$

$\theta_{min} = 1.22\times 9.145 \times 10^{-6}$

$\theta_{min} = 1.21 \times 10^{-5}\ rad$

the minimum angle of spreading is $\theta_{min} = 1.21 \times 10^{-5}\ rad$

5 0

2 years ago

Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo

zubka84 [21]

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

3 0

2 years ago

A projectile follows a straight-line path instead of a parabolic trajectory. Which could be the launch angle? would it be 90 or

Gnesinka [82]

<em>projectile can only follow the straight line path when it is launched upward straightly so the correct option is <u>90 degree with respect to horizontal x -axis ..:)</u></em>

7 0

2 years ago

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 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$