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3 years ago

How are the electric field lines around a positive charge affected when a second positive charge is near it?

Physics

1 answer:

Liono4ka [1.6K]3 years ago

8 0

Answer:Because it likes to repel and attract opposites

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A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the

Ierofanga [76]

Answer:

The discharge rate is $Q = 0.0192 \ m^3 /s$

Explanation:

From the question we are told that

The diameter is $d = 60 \ mm = 0.06 \ m$

The head is $h = 6 \ m$

The coefficient of contraction is $Cc = 0.68$

The coefficient of velocity is $Cv = 0.92$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{ 0.06 }{2}$

$r = 0.03 \ m$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.03)^2$

$A = 0.00283 \ m^2$

The discharge rate is mathematically represented as

$Q = Cv *Cc * A * \sqrt{ 2 * g * h}$

substituting values

$Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}$

$Q = 0.0192 \ m^3 /s$

6 0

2 years ago

What is the average power supplied by a 60.0 kg secretary running up a flight of stairs rising vertically 4.0 m in 4.2 s?

gayaneshka [121]

Answer:

9.8kW

Explanation:

Given data

Mass= 60kg

Hieght= 4m

Time= 4.2seconds

We know that the energy possessed is given as

PE=mgh

PE=60*9.81*4

PE= 2354.4 Joulse

Also, the expression for power is

Power=Energy*Time

Power= 2354.4*4.2

Power=9888.48 watt

Power= 9.8kW

4 0

3 years ago

Which statements about electric field lines are correct? Check all that apply.

slava [35]

Answer:

they cross over one another between charge.

7 0

3 years ago

How big is omniverse?

klio [65]

Answer:

big half like whale is so big

8 0

2 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago