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3 years ago

A 20-ton truck collides with a 1500-lb car and causes a lot of damage to the car. During the collision:

Physics

1 answer:

dexar [7]3 years ago

6 0

Answer:

B. the force of on the truck due to the collision is exactly equal to the force on the car.

Explanation:

At the time of collision between two objects of unequal or equal masses two forces emerge at the point of collision .They are called action and reaction force . According to third law of Newton , they are equal . So in the instant case of collision between car and truck , force on both of them will be equal.

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A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

My name is Ann [436]

Answer:

a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C , c) U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

C = k e₀ A / d

Let's calculate

C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

C = 4,012 10⁻¹⁴ F

b) let's look the charge

C = Q / ΔV

Q = C ΔV

Q = 4,012 10⁻¹⁴ 400

Q = 1.6 10⁻¹¹ C

c) The stored energy

U = ½ C ΔV²

U = ½ 4,012 10⁻¹⁴ 400²

U = 3.21 10⁻¹¹ J

4 0

3 years ago

Noise-canceling headphones are an application of destructive interference. Each side of the headphones uses a microphone to pick

exis [7]

The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.

4 0

3 years ago

I wonder if people that say that there ugly is to get attention (not to be mean...)

alexandr1967 [171]

They do. mostly because they want validation from others

5 0

2 years ago

Read 2 more answers

An object is 2.0 m in front of a plane mirror. Its image is:?

maria [59]

Planes don’t have mirrors

6 0

3 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago