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mr_godi [17]
4 years ago
9

2.)

Physics
1 answer:
Oduvanchick [21]4 years ago
4 0

Answer:

-22.7 m/s^2

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2

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Recall what you learned about the state of Virginia. Choose the statements that describe the paleontology of the
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An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by p
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Complete Question

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?

A It should stay the same

B  It should be quadrupled.

C It should be quintupled

D It should be doubled.

E It should be tripled

Answer:

Option D is the correct option

Explanation:

Generally electric field is mathematically represented as

        E =  \frac{\sigma}{\epsilon_o}

Where \sigma is the charge per unit area (Charge density )

From the question we are told that \sigma is doubled hence the

     E =  \frac{2 \sigma }{\epsilon_o}    

Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density

8 0
3 years ago
A uniform, solid cylinder of radius 5.00 cm and mass 3.00 kg starts from rest at the top of an inclined plane that is 2.00 m lon
ch4aika [34]

To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is

KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f

KE_i = 0, since initially cylinder was at rest

PE_f = 0 since at the ground potential energy is zero

The mathematical values are,

mgh = \frac{1}{2} mV^2 + \frac{1}{2}I\omega^2

Here,

m = mass

g= Gravity

h = Height

V = Velocity

I = \frac{mr^2}{2} moment of Inertia in terms of its mass and radius

\omega = \frac{V}{r} Angular velocity in terms of tangential velocity and its radius

Replacing the values we have that

mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2

gh = \frac{v^2}{2}+\frac{v^2}{4}

v = \sqrt{\frac{4gh}{3}}

From trigonometry the vertical height of inclined plane is the length of this plane for sin\theta, then

h = 2.00*sin 25

h = 0.845 m

Replacing,

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V = 3.32 m/s

Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s

6 0
3 years ago
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