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3 years ago

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a block of wood with a mass of 2.33 kg is at rest on a frictionless pole. A .011 kg bullet is fired 722 m/s and is embedded with

the wood. What is the speed of the bullet block system after the collision occurs.

1 answer:

MrRissso [65]3 years ago

6 0

Some rewards are 2.33 miles in a hour so you have to move in 700 degrees to get the system moving faster soo 700+ 2.33 divide by 3

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Weight because both gravity and weight are pulling an object down

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How do you determine the acceleration of an object?

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Are you familiar with any basic calculus? If so, we can just look at this derivative and see what's happening with our units here..

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3 years ago

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Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl

tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

$v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s$

For Train 2 Velocity of the Source,

$v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s$

Frequency of Source,

$f_{s}=500\ Hz$

To Find:

Frequency of Observer,

$f_{o}=?$ (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source $v_{s}$ and observer $v_{o}$ , the original frequency of the sound waves $f_{s}$ and the observed frequency of the sound waves $f_{o}$ ,

The Equation is

$f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})$

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

$f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz$

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

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3 years ago

The equation below can be used to find the specific heat capacity of a substance. What is the specific

slavikrds [6]

Answer:

<em>specific</em><em> </em><em>heat</em><em> </em><em>capacity</em><em> </em><em>(</em><em>c</em><em>)</em><em> </em><em>=</em>107.2J/kg K

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2 years ago

A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele

ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

$u^2-u'^2 = 2as$

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

$u^2-0^2 = 2\times9.8\tiimes1.08$

or

$u = \sqrt{21.168}$

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed ( $v_f$ ) = v = 4.6 m/s

Then conservation of energy says

$\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2$

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s

u₂ = velocity of the basketball before collision= 4.6 m/s

v₁ = velocity of the tennis ball after collision

v₂ = velocity of the basketball after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

$v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2$

substituting the values in the above equation we get

$v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6$

or

$v = -8.565-3.965$

or

$v = -12.53m/s$

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

$K.E = \frac{1}{2}mv^2$

or

$K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2$

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

5 0

3 years ago