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1 year ago

In which direction does the electric field point at a position directly east of a positive charge

Physics

2 answers:

lora16 [44]1 year ago

7 0

Field lines always point away from the positive side of a magnet. So i would say east but im not to sure

natima [27]1 year ago

5 0

Answer: Towards East

Explanation:

The direction of the electric force per unit charge i.e. the electric field is given by the direction of motion of positive test charge under the electric force. The field lines are directed radially outwards for a positive charge and radially inwards for a negative charge.

Thus, the electric field points towards East for the position directly east of a positive charge

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an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and

natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\\\\\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s$

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.

4 0

1 year ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

ipn [44]

Answer:

ball clears the net

Explanation:

$v_{o}$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_{ox}$ = initial velocity = $v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}$ Eq-1

Consider the motion of the ball along the vertical direction

$v_{oy}$ = initial velocity = $v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}$

$t$ = time of travel

$Y_{o}$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_{y}$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\$

Since Y > 1 m

hence the ball clears the net

7 0

2 years ago

The information about groundwater recorded by a student is valid when

skelet666 [1.2K]

Such information is valid when A. it is not biased.
There is no bias in science - the results are either correct or incorrect; personal opinions have nothing to do with that. B is incorrect because the point of experiments is that they can be repeated to achieve the same results. C and D are likewise incorrect as primary sources are important, and you can share the results with others if you want to.

7 0

2 years ago

Read 2 more answers

Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through the material of

gtnhenbr [62]

Answer:

y_red / y_blue = 1.11

Explanation:

Let's use the constructor equation to find the image for each wavelength

1 /f = 1 /o + 1 /i

Where f is the focal length, or the distance to the object and i the distance to the image

Red light

1 / i = 1 / f - 1 / o

1 / i_red = 1 / f_red - 1 / o

1 / i_red = 1 / 19.57 - 1/30

1 / i_red = 1,776 10-2

i_red = 56.29 cm

Blue light

1 / i_blue = 1 / f_blue - 1 / o

1 / i_blue = 1 / 18.87 - 1/30

1 / i_blue = 1,966 10-2

i_blue = 50.863 cm

Now let's use the magnification ratio

m = y ’/ h = - i / o

y ’= - h i / o

Red Light

y_red ’= - 5 56.29 / 30

y_red ’= - 9.3816 cm

Light blue

y_blue ’= 5 50,863 / 30

y_blue ’= - 8.47716 cm

The ratio of the height of the two images is

y_red ’/ y_blue’ = 9.3816 / 8.47716

y_red / y_blue = 1,107

y_red / y_blue = 1.11

5 0

2 years ago

Your film idea is about drones that take over the world. In the script, two drones are flying horizontally at the same speed and

Stella [2.4K]

Answer:

vₓ = 20 m/s, v_{y} = -15 m / s

Explanation:

This is a conservation of moment problem, since it is a vector quantity we can work each axis independently

The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved

X axis

Initial moment. Before the crash

p₀ = m₁ v₀ₓ + m₂ v₀ₓ

Final moment. After the crash

p_{fx} = (m₁ + m₂) vₓ

p₀ₓ = $p_{fx}$

m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ

vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)

vₓ = v₀ₓ = 20 m/s

Y Axis

Initial

p_{oy} = m₁ v_{oy}

Final

p_{fy} = (m₁ + m₂) v_{y}

p_{oy} = p_{fy}

the drom rises and when it falls it has the same speed because there is no friction v_{oy} = -60 m/s

m₁ $v_{oy}$ = (m₁ + m₂) v_{y}

v_{y} = m₁ / (m₁ + m₂) v_{oy}

v_{y} = 1/4 60

v_{y} = -15 m / s

Vertical speed is down

5 0

2 years ago