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3 years ago

Ball A (mass of 0.5 kg) going +2.0 m/s collides with a stationary ball B (mass of 0.4 kg).

Physics

1 answer:

ki77a [65]3 years ago

6 0

The correct answer is 1.2 m/s

: mv+mv=mv+mv

(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)

= 1kg*m/s=(0.5kg)v+0.4kg*m/s

=1kg*m/s-0.4kg*m/s=(0.5kg)v

=0.6kg*m/s=(0.5kg)v

to solve for v we divide both side by 0.5kg

v=1.2m/s

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That's called the wave's "wavelength" .

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Suppose you have a rock that, when it solidifies, contains 1 microgram of a radioactive isotope. How much of this isotope remain

algol13

Answer:

d) 1/32 microgram

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Forth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

Fifth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/32.

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3 years ago

You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building.

DanielleElmas [232]

Answer:

t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

v = v₀ + a t₁

the elevator starts with zero speed

v = a t₁

t₁ = v / a

t₁ = 18.8 / 2.40

t₁ = 7.833 s

in this time he runs

y₁ = v₀ t₁ + ½ a t₁²

y₁ = ½ a t₁²

y₁ = ½ 2.40 7.833²

y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

x_total = x₁ + x₂

x₂ = x_total - x₁

x₂ = 373 - 73,627

x₂ = 299.373 m

this distance travels at constant speed,

v = x₂ / t₂

t₂ = x₂ / v

t₂ = 299.373 / 18.8

t₂ = 15.92 s

therefore the total travel time is

t_total = t₁ + t₂

t_total = 7.833 + 15.92

t_total = 23.757 s

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3 years ago

Placed exactly between two oppositely charged point charges, a test charge (the

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Zero maximum force (N) or field strength (N/C). ... minimum /maximum field strength.

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3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago