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2 years ago

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a bullet of mass m is fired into a block of mass m that is at rest. the block, with the bullet embedded, slides distance d acros

s a horizontal surface. the coefficient of kinetic friction is μk.

2 answers:

AVprozaik [17]2 years ago

7 0

Uk=2m*the normal force
the distance d is needed only if you were asked about work

Ilya [14]2 years ago

6 0

The expression for the bullet's speed vbullet is $v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}$ and the speed of a 9.0 g bullet is v=16.8 m/s

<h3>Explanation: </h3>

A bullet of mass m is fired into a block of mass m that is at rest. the block, with the bullet embedded, slides distance d across a horizontal surface. the coefficient of kinetic friction is μk.

The initial kinetic energy of the bullet is given by

$K_i = \frac{1}{2}mv^2$

where

m is the mass of the bullet

v is the initial speed of the bullet

The expression for the bullet's speed vbullet is

$F=\mu_k (m+M) g$

where

$\mu_k$ is the coefficient of kinetic friction

g is the acceleration of gravity

The work done by the force of friction is

$W=-Fd = -\mu_k (m+M)g d$

where d is the displacement of the block+bullet.

Because the final kinetic energy is zero (the bullet with the block comes at rest), we can write:

$W=K_f - K_i = -K_i$

And so

$-\mu_k (m+M) g d = -\frac{1}{2}mv^2$

By solving for v, the solution for the bullet speed:

$-\mu_k (m+M) g d = -\frac{1}{2}mv^2\\v=\sqrt{\frac{2\mu_k (m+M)gd}{m}}$

The speed of a 9.0 g bullet that, when fired into a 12 kg stationary wood block causes the block to slide 5.4 cm across a wood table. Assume that k=0.20.

We have:

the mass of the bullet, m = 9.0 g = 0.009 kg

the mass of the block, M = 12 kg

the distance covered by the block+bullet, d = 5.4 cm = 0.054 m

the coefficient of friction, $\mu_k = 0.20$

the acceleration of gravity, $g = 9.8 m/s^2$

By substituting, we got

$v=\sqrt{\frac{2 (0.20) (0.009 kg+12 kg)(9.8 m/s^2)(0.054 m)}{0.009 kg}}=16.8 m/s$

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A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

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2 years ago

a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll

marissa [1.9K]

Answer:

$h = 0.0362\,m$

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

$U_{g,A} + K_{A} = U_{g,B} + K_{B}$

$U_{g,B} - U_{g,A} = K_{A} - K_{B}$

$m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}$

The moment of inertia of the coin is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

After some algebraic handling, an expression for the maximum vertical height is derived:

$m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}$

$h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}$

$h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 0.0362\,m$

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2 years ago

A cart, which has a mass of 2.30 kg is sitting at the top of an inclined plane, which is 4.50 meters long and meets the horizont

expeople1 [14]

Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) The magnitude of the force that causes the cart to roll down is 5.47 N.

c) The acceleration of the cart is 2.38 m/s²

d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

Where:

EP = gravitational potential energy.

m = mass of the object.

g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

4.50 m · sin 14.0° = height

height = 1.09 m

Then, the gravitational potential energy will be:

EP = m · g · h

EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J

The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

Fwx = 5.47 N

The magnitude of the force that causes the cart to roll down is 5.47 N.

c)Using the equation of Fwx we can calculate the acceleration of the cart:

Fwx = m · a

Where "m" is the mass of the cart and "a" is the acceleration.

Fwx / m = a

5.47 N / 2.30 kg = a

a = 2.38 m/s²

The acceleration of the cart is 2.38 m/s²

d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:

x = 1/2 · a · t²

4.50 m = 1/2 · 2.38 m/s² · t²

2 · 4.50 m / 2.38 m/s² = t²

t = 1.94 s

It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:

v = v0 + a · t

v = 0 m/s + 2.38 m/s² · 1.94 s

v = 4.62 m/s

The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the cart.

v = velocity of the cart.

KE = 1/2 · 2.30 kg · (4.62 m/s)²

KE = 24.6 J

The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

The gain of kinetic energy is equal to the loss of gravitational potential energy.

g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:

W = -ΔPE

W = -(final potential energy - initial potential energy)

W = -(0 - 24.6 J)

W = 24.6 J

This can also be calculated using the definition of work:

W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

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2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

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2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

b)

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

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2 years ago