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alex41 [277]
3 years ago
6

Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)

, and the second charge has magnitude of q2=3.60 nCq2=3.60 nC and is located at (x=1.20 m,y=0.600 m)(x=1.20 m,y=0.600 m), calculate the xx and yy components, ExEx and EyEy, of the electric field, E⃗ E→, in component form at the origin, (0,0)(0,0). The Coulomb force constant is 1/(4π????0)=8.99×109 N⋅m2/C21/(4πϵ0)=8.99×109 N⋅m2/C2.
Physics
1 answer:
Gala2k [10]3 years ago
8 0

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

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3 years ago
A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang
koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × 10^{-4}  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = \int \vec{B}.d\vec{A}   .................1

and Ф = BA cos∅    ............2

so B = \frac{\phi }{Acos\theta }    

and we know

A = ab

so

B = \frac{\phi }{abcos\theta }    ..............3

put here value

B =  \frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}  

solve we get

B = 0.692 T

8 0
3 years ago
Children in a tree house lift a small dog in a basket 4.00 m up to their house. If it takes 187 J of work to do this, what is th
Vika [28.1K]
A=Fh
A - work
F - force
h - distance

F=mg
m - mass (god+basket)

so
A=mgh
187 = m*10*4
187=40m
m=187/40
m=4.675 kg
or 4kg and 675g

pretty small dog...
6 0
3 years ago
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