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goblinko [34]
3 years ago
10

A long cylindrical insulator has a uniform charge density of 1.54 µc/m3 and a radius of 6cm what is the electric field inside th

e insulator at a distance of 2 cm

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0
Plug given values in SI units to get electric field.

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Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45
victus00 [196]
<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>
4 0
3 years ago
When you park your car uphill next to a curb, the right front wheel should be:?
Cloud [144]
<span> When headed uphill at a </span>curb<span>, turn the front </span>wheels<span> away from the </span>curb<span> and let </span>your vehicle<span> roll backwards slowly until the rear part of the front </span>wheel<span> rests against the </span>curb<span> using it as a block.</span>
6 0
3 years ago
In which type of bone does ossification occur in the membranes?
Afina-wow [57]
In the formation of flat bones such as the skull the mandibles and the clavicles  
3 0
3 years ago
A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 12
PSYCHO15rus [73]

Answer: 6.36

Explanation:

Given

Radius of grindstone, r = 4 m

Initial angular speed of grindstone, w(i) = 8 rad/s

Final angular speed of the grindstone, w(f) = 12 rad/s

Time used, t = 4 s

Angular acceleration of the grinder,

α = Δw / t

α = w(f) - w(i) / t

α = (12 - 8) / 4

α = 4/4 = 1 rad/s²

Number of complete revolution in 4s =

Δθ = w(i).t + 1/2.α.t²

Δθ = 8 * 4 + 1/2 * 1 * 4²

Δθ = 32 + 1/2 * 16

Δθ = 32 + 8

Δθ = 40 rad/s

40 rad/s = 40/2π rpm = 6.36 rpm

Therefore, the grindstone does 6.36 revolutions during the 4 s interval

6 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

a = -\omega^2 A

Where,

a = Acceleration

A = Amplitude

\omega= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

a = -g

-\omega^2 A = -g

\omega = \sqrt{\frac{g}{A}}

From the definition of frequency and angular velocity we have to

\omega = 2\pi f

f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}

f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}

f = 2.5Hz

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0
3 years ago
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