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2 years ago

Hi! Can somebody please help?

Physics

1 answer:

dezoksy [38]2 years ago

7 0

Answer:

Diagram A will reach the top first.

Explanation:

If it is going straight, it will go slower. The higher the movement speed the faster it is. Hope this helps!

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Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

Calculate the acceleration of a train travelling from rest to 24 m/s in 12 seconds.

Vilka [71]

Answer: A 2m/s^2

Steps: Formula for acceleration. (Velocity Final - Initial Velocity) / Time

(24 - 0) / 12 = 2

3 0

2 years ago

HELP ASAP <br> Describe when contact metamorphism occurs?

diamong [38]

Contact metamorphism occurs adjacent to igneous intrusions and results from high temperatures associated with the igneous intrusion. Since only a small area surrounding the intrusion is heated by the magma, metamorphism is restricted to the zone surrounding the intrusion, called a metamorphic or contact aureole

5 0

2 years ago

Read 2 more answers

A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?

Nadusha1986 [10]

Answer:

0m/s

Explanation:

Since its fired at an angle, at the top there will be a split second where the velocity will be 0, as it has a parabolic shape, so the speed at the top of its path is 0

4 0

2 years ago

A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in

melamori03 [73]

Answers:

a) $0.5 m/s^{2}$

b) $1.5 N$

Explanation:

a) The centripetal acceleration $a_{c}$ of an object moving in a uniform circular motion is given by the following equation:

$a_{c}=\omega^{2} r$

Where:

$\omega=1 \frac{rev}{s}$ is the angular velocity of the ball

$r=0.5 m$ is the radius of the circular motion, which is equal to the length of the string

Then:

$a_{c}=(1 \frac{rev}{s})^{2} 0.5 m$

$a_{c}=0.5 m/s^{2}$ This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force $F$ ) that is directed towards the center and is equal to the tension ( $T$ ) in the string:

$F=T=m. a_{c}$

Where $m=3 kg$ is the mass of the ball

Hence:

$T=(3 kg)(0.5 m/s^{2})$

$T=1.5 N$ This is the tension in the string

7 0

2 years ago