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2 years ago

Determine the electrical force of attraction between two balloons

Physics

1 answer:

Deffense [45]2 years ago

8 0

Force = 9x10^9 x (6x10^-7 x 6x 10^-7) / (.5)^2
= 0.013 N

Using the formula;

F = 9x10^9 x (Q1 x Q2) / r^2

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A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.

gogolik [260]

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Square root of (4^2 + 4^2) = 4*squareRoot(2)

7 0

2 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago

All pressure topics in physics

AfilCa [17]

Absolute, Atmospheric, Differential, and Gauge Pressure

5 0

1 year ago

Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of

kvv77 [185]

Answer: $V_{net}=4\frac{kQ}{a}$

Explanation:

Given

charge on each Particle is Q

Length of diagonal of the square is 2a

therefore distance between center and each charge is $\frac{2a}{2}=a$

Electric Potential of charged Particle is given by

For First Charge

$V_1=\frac{kQ}{a}$

$V_2=\frac{kQ}{a}$

$V_3=\frac{kQ}{a}$

$V_4=\frac{kQ}{a}$

total Electric Potential At center is given by

$V_{net}=V_1+V_2+V_3+V_4$

$V_{net}=4\times \frac{kQ}{a}$

$V_{net}=4\frac{kQ}{a}$

7 0

2 years ago

PLEASE HELP ASAP

77julia77 [94]

Answer:

<em>The first choice (32m/s) is the closest to the answer</em>

Explanation:

The magnitude of a vector is the distance between the initial and the end point of the vector.

Being Vx and Vy the horizontal and vertical components of the vector V respectively, the magnitude of V is calculated as:

$\mid \vec{V} \mid =\sqrt{V_x^2+V_y^2}$

The components of the velocity of the physics student's projectile launcher are Vx=28 m/s and Vy=15 m/s.

Calculate the magnitude of the velocity:

$\mid \vec{V} \mid =\sqrt{28^2+15^2}$

$\mid \vec{V} \mid =\sqrt{784+225}$

$\mid \vec{V} \mid =\sqrt{1009}$

$\mid \vec{V} \mid =31.8 \ m/s$

The first choice (32m/s) is the closest to the answer

8 0

2 years ago