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4 years ago

A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.

1 answer:

4 0

The acceleration of the object which moves from an initial step to a full halt given the distance traveled can be calculated through the equation,
d = v² / 2a
where d is distance, v is the velocity, and a is acceleration
Substituting the known values,
180 = (22.2 m/s)² / 2(a)
The value of a is equal to 1.369 m/s²
The force needed for the object to be stopped is equal to the product of the mass and the acceleration.
F = (1300 kg)(1.369 m/s²)
F = 1779.7 N

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Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

d sin θ= m λ constructive interference

d sin θ = (m + ½) λ destructive interference

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tanθ = y / L

tan te = sin tea / cos tea = sin tea

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derestimate the first two equations.

Let's do the calculation for constructive interference

d y / L = m λ

the distance between maximum clos is and

y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

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y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

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3 years ago

A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on

klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, $q=1.60218\times 10^{-19}\ C$

Mass of a proton, $m=1.67262 \times 10^{-27}\ kg$

The cyclotron frequency is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}$

f = 2286785.40 Hz

$\omega=14368296.44\ rad/s$

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2) They can travel in a vacuum = both

3) They have energy = both

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