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1 year ago

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A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself.

The bullet-block system slides forward 2.4 m before coming to rest.

1 answer:

aksik [14]1 year ago

7 0

What is the coefficient of kinetic friction between the block and the surface? Express your answer using two significant figures.

Answer:

0.39

Explanation:

Given information

m1=1.7 Kg

m2=0.011 Kg

v2=670 m/s

d=2.4 m

m2v2=(m1+m2)v hence $v=\frac {m2v2}{m1+m2}$ and also $a=\frac {v^{2}}{2d}$

The deceleration due to friction is given by

$F=\mu_k N=\mu_k W=\mu_k (m1+m2)g$

$F=(m1+m2)a=\mu_k (m1+m2)g$ therefore $a=\mu_k g$

Therefore, $a=\frac {v^{2}}{2d}=\mu_k g$

$\mu_k=\frac {1}{2dg}(\frac {m2v}{m1+m2})^{2}=\frac {1}{2*2.4*9.81}\times (\frac {0.011*670}{1.7+0.011})^{2}$

$\mu_k=0.39$

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A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an

Sonbull [250]

Answer:

(a) k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = $\frac{M}{3} (a^{2} +b^{2} )$ ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k = $\frac{1}{2} Iw^{2}$ --------------------------------2

Putting equation 1 into equation 2, we have;

k = $\frac{M}{6} (a^{2} +b^{2} )w^{2}$

k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = $\frac{Mw}{3} (a^{2} +b^{2} )$

But

τ = dτ/dt = $\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}$ ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

8 0

2 years ago

A mover pushes a 255 kg piano

faust18 [17]

Answer:

$0.495 ms^{-2}$

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

$F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}$

7 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. The heat capacity of Object A is

Oksi-84 [34.3K]

Answer:

Explanation:

Heat capacity A = 3 x heat capacity of B

initial temperature of A = 2 x initial temperature of B

TA = 2 TB

Let T be the final temperature of the system

Heat lost by A is equal to the heat gained by B

mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)

heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)

3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)

3 TA - 3 T = T - TB

6 TB + TB = 4 T

T = 1.75 TB

8 0

2 years ago

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance

Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

so here speed ratio is

initial speed = $\frac{40}{25}$

so initial speed = 1.6

so

stopping distance increase = (1.6)²

$\frac{D40}{D25}$ = (1.6)²

$\frac{D40}{D25}$ = 2.56

so here

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

5 0

2 years ago