65mi/hr South is an example of

Find the magnitude of the electric field at a point midway between two charges +13.6 x 10-9 C and +61.0 x 109 C separated by a d

Ostrovityanka [42]

Answer:

Ep = 3797.05 N/C in the direction leaving the charge q₂ towards point P

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Data

k= 8.99*10⁹ N*m²/C²

q₁ = +13.6 x 10⁻⁹C

q₂ = +61.0*10⁻⁹C

d₁ =d₂= 33.5 cm = 0.335 m

Look at the attached graphic:

E₁: Electric Field at point P due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge

E₂: Electric Field at point P due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 8.99*10⁹ *13.6 *10⁻⁹/(0.335)² = 1089.45 N/C

E₂ = k*q₂/d₂²=- 8.99*10⁹ *61*10⁻⁹/(0.335)² = - 4886.5 N/C

Magnitude of the electric field at a point midway between q₁ and q₂

Ep= - 4886.5+ 1089.45 = -3797.05 N/C

Ep = 3797.5 N/C in the direction leaving the charge q₂ towards point P

6 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and

vladimir1956 [14]

Answer:

a) t= 0.92 s

b) h = 0.46 m

c) v = -6.04 m/s

Explanation:

A)

In order to find the total time that the feet are in the air, we must add two times:
1) time needed to reach to the maximum height (t₁)
2) time from when starts to fall from the maximum height until her feet hit the water (t₂)
In order to get t₁, we need to take into account that at her highest point, the vertical speed will be zero.
Taking for granted the value for the acceleration due to gravity,

g = -9.8 m/s2, we can apply the definition of acceleration, and replacing by the givens, we can find t₁ as follows:

$t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)$

In order to find t₂, we need to find first the highest point above the board, which is indeed what is asked for in b).
We can use the following kinematic equation, taking into account that at the highest point, the final velocity vf will be zero.
The equation can be written as follows:

$v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h (2)$

Replacing by the givens, and solving for Δh, we get:

$\Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)$

The total height over the water will be just the sum of the takeoff point (1.40 m over the water) and the value we found in (3):
H = 1.40 m + 0.46 m = 1.86 m
Now, we can use the equation that relates the vertical displacement with the time, remembering that v₀=0, as follows:

$H = \frac{1}{2}*g*t^{2} (4)$

Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:

$t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)$

The total time will the sum of t₁ and t₂:
t = 0.3 s + 0.62 s = 0.92 s (6)

B)

As we have just found, the highest point above the board was at 0.46m above the takeoff point, so the highest point above the board is just 0.46 m.

C)

In order to find the velocity when her feet hit the water, we can use the same equation (2), taking into account that v₀=0, and Δh = H= -1.86m.
Solving (2) for vf, we get:

$v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)$

5 0

3 years ago

Suppose the sphere is positively charged. Is it attracted to, repelled by, or not affected by the magnet?.

Nutka1998 [239]

When a positively charged sphere is brought near the north pole of a magnet, the positively charged sphere will be attracted to the magnet.

<h3>Positively charged object</h3>

When a positively charged object is brought near the north pole of a magnet, the positively charged object will be attracted to the magnet beacuse of polarity.

Positively charged metals have the tendency to show the polarization of charges.

Thus, when a positively charged sphere is brought near the north pole of a magnet, the positively charged sphere will be attracted to the magnet. Also, if the south pole is brought near the sphere, the positively charged sphere will repel the magnet.

Learn more about attraction of magnet here: brainly.com/question/14749231

4 0

3 years ago

An equipotential surface that surrounds a point charge q has a potential of 536 V and an area of 1.20 m2. Determine q.

gladu [14]

Answer:

q = 1.84×10^-8coulombs

Explanation:

Surface area = 4πr²

r is the distance

1.2 = 4(3.14)r²

1.2 = 12.56r²

r² = 1.2/12.56

r² = 0.0956

r = √0.0956

r = 0.309m

Get the charge C

V = kq/r

536 = 9.0×10^9q/0.309

536×0.309 = 9×10^9q

165.73 = 9×10^9q.

q = 165.73/9×10^9

q = 1.84×10^-8coulombs

4 0

3 years ago

The heat Fusion is the amount of heat required to__

Annette [7]

C. Melt 1g if solid into liquid.

5 0

4 years ago

65mi/hr South is an example of​

65mi/hr South is an example of