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2 years ago

What are the magnitude and direction of electric field at the center of square?

Physics

1 answer:

Kobotan [32]2 years ago

6 0

Answer:

hello the diagram related to this question is missing attached below is the missing diagram

Answer :

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

Explanation:

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

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a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration

slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0

2 years ago

A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori

Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²
 That's about 2.634 m/s² 

(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = 2.45 N.

That's the force that's accelerating the little block, so that must be the tension
in the string.

7 0

3 years ago

The eficiency of a simple machine can never be 100% why?

Kitty [74]

Answer:

Systems always tend toward a state of decreasing order unless more energy is provided into the system to counteract this tendency.

6 0

3 years ago

A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe

liubo4ka [24]

Answer:

ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

$I_1=\dfrac{M+m}{2}r^2$

$I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2$

$I_1=0.75\ kg.m^2$

Final mass moment of inertia

$I_2=\dfrac{M}{2}r^2$

$I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2$

$I_2=0.468\ kg.m^2$

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

$\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s$

ω₂=1.20

7 0

3 years ago

A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler

KIM [24]

Answer:

(a) m = 33.3 kg

(b) d = 150 m

Explanation:

Newton's second law to the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Data

F= 100 N

a= 3.0 m/s²

(a) Calculating of the mass of the block:

We replace dta in the formula (1)

F = m*a

100 = m*3

m = 100 / 3

m = 33.3 kg

Kinematic analysis

Because the block moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t Formula (3)

Where:

d:displacement in meters (m)

t : time interval in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

a= 3.0 m/s²

v₀= 0

t = 10 s

(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

vf= 30 m/s

4 0

3 years ago