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skad [1K]
3 years ago
10

A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t

he test charge? (k = 9.0 × 109 newton·meter2/coulomb2)
Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance. 

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C
Sveta_85 [38]3 years ago
8 0

Answer:

E.  9.2 × 108 newtons/coulomb.

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A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s
TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

vf^{2} =vo^{2}+2*g*h\\

We clear the vo (initial speed)

vo=\sqrt{vf^{2}-2*g*h }

v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}

vo=5.87m/s

7 0
2 years ago
Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.
jasenka [17]

(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

<h3>Force per unit length</h3>

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d -  I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L =  (μI/2π) x (I/2d + I/d)

F₃/L =  (μI/2π) x (3I/2d)

F₃/L =  3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

Learn more about magnetic force here: brainly.com/question/13277365

#SPJ1

6 0
2 years ago
Please answer this question
sergij07 [2.7K]

Explanation:

m = kg. v=m/s. g=m/s^2. h= m

>>1/2mv^2=mgh

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .

<u>>>kg m^2/s^2=kg m^2/s^2</u><u> </u>

<u>></u><u>></u><u>J</u>= J

3 0
3 years ago
A car is driven 125.0 km due west then 65.0 km due south. What is the magnitude of its displacement?
kodGreya [7K]

pythagoras' theorem on right angled triangle. sides 125, 65

sqrt (125^2 +65^2)

5 0
3 years ago
Which best describes the image of a concave mirror when
mr_godi [17]

Answer:

when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted

Explanation:

Let's use the constructor equation to describe the image of a concave mirror

    1 / f = 1 / p + 1q

where f is the focal length, p and q the distance to the object and the image, respectively

     1 /q = 1/f - 1/p

tell us that the image is between the focal and twice the focal, let's calculate the position of the image

for both ends

case 1, distance to the object       p = f

     

       1 / q = 1 / f -1 / f

       1 / q = 0

       q = ∞

    the image is in infinity

case2, distance to object p = 2f

     1 / q = 1 / f - 1 / 2f

     1 / q = 1 / 2f

     q = 2f

the image is twice the focal length, the object and the image are at the same point

therefore the image when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted

5 0
2 years ago
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