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3 years ago

13

Divers in acapulco, mexico, dive headfirst at 8 feet per second from the top of a cliff 87 feet above the pacific ocean. during

which time period will the diver's height exceed that of the cliff?

1 answer:

xxMikexx [17]3 years ago

8 0

Depends on the wieght of his genitals.

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At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago

An unknown additional charge q3 is now placed at point B, located at coordinates (0 m, 15.0 m ). Find the magnitude and sign of

Finger [1]

The first part of the question is:

Two point charges are placed on the x axis. (Attached file)The first charge, q1 = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Answer:

q3 = +0.3nc

Explanation:

Due to the vector symbols in the solution, I've decided to attach the explanation to this answer.

5 0

3 years ago

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Savatey [412]

We are given the equation:

<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s ) = 56.65 m/s

</span>For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24<span> meters) / (2.40 s - 2.20 s ) = 50.6 m/s</span>

4 0

3 years ago

A car travelling at 10 m/s accelerates at 3 m/s^2 for 6 seconds. What is the final velocity of the car?

tankabanditka [31]

Answer:28m/s

Explanation:I got one point so yeah hope this help

7 0

3 years ago

A motorcycle starts to move from rest. If the velocity of the motorcycle becomes 90 km/hr

bagirrra123 [75]

Explanation:

90 kmhr—1 x 1000/3600 = 25ms—1

U = 0 ms—1

V = 25ms—1

t = 10 s

a = ?

a = V - U/t

a = 25 - 0/10

a = 25/10

a = 2.5 ms—1

6 0

2 years ago