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KonstantinChe [14]

3 years ago

13

A ballistics test involves firing a 10.0g bullet horizontally into a 1.50kg wood block that is initially at rest. The bullet bec

omes lodged in the block, and you determine that they move together at a speed of 5.00m/s. With this information, determine the velocity of the bullet before it entered the block.

1 answer:

wariber [46]3 years ago

4 0

Answer:

u= 755 m/s

Explanation:

Lets take

Velocity of bullet = u m/s

The velocity of system after collision ,v= 5 m/s

m= 10 g = 0.01 kg

M= 1.5 kg

From linear momentum conservation( because there is no any external force on the system)

Pi= Pf

m u = ( m +m) v

0.01 x u= (1.5 + 0.01) x 5

u= 755 m/s

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A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl

Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

The velocity field with which the bird is flying is $\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

The temperature of the room is $T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C$

The time considered is t = 10 \ seconds

The distance that the bird flew is x = 1 m

Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

$\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

So

$\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]$

From the given equation of velocity field

$v_x = 0.6x$

$v_y = 0.2t$

$v_z = -1.4$

So

$\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]$

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]$

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$

$\frac{dT}{dt} = 1.016 ^oC/m$

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Select all that apply. The resistance to an object being pulled over a surface is called _____. starting friction, sliding frict

jok3333 [9.3K]

That is kinetic friction.

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If the current flowing through an electric heater increases from 6 to 12 amp while the voltage remains the same, the heat produc

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Q=l^2 times R times t
Where Q - heat, I -current, R - resistance and t is time

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4 years ago

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A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$

The maximum force is when the car is accelerating so

$Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N$

so the maximum force is the maximum force by the maximum speed

$Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW$

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

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The law of reflection states that the angle of reflection is equal to the angle of

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The law of reflection states that the angle of reflection is equal to the angle of Incidence .

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