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Harlamova29_29 [7]
3 years ago
5

The strength of the electric field 0.5 m from a 6 uC charge is

Physics
1 answer:
Triss [41]3 years ago
5 0

Answer:

E= 2.158× 10*5N/C

Explanation:

K= 8.99×10*9, q= 6×10*-6C, d= 0.5m

E= kq/d*2

E= (8.99×10*9× 6×10*-6)/0.5*2

E= 215760

E= 2.158 ×10*5N/C

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
Astronomers define the __________ as all of space and everything in it. It is enormous, almost beyond imagination. Question 2 op
Genrish500 [490]

Answer:

Universe

Explanation:

I took the quiz.

3 0
3 years ago
Which qualifications are typical for a Manufacturing career? Check all that apply.
kicyunya [14]
D should be the answer
8 0
3 years ago
Why do astronauts need to wear pressurized suits in space?
SashulF [63]
The correct answer is C , because the space is vacuum and his body can explode and for this reason,  the astronaut need a special costum to be protected. It's the same on the moon, because there is no atmosphere 
6 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
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