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3 years ago

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A shell traveling with speed, v0 , exactly horizontally and due north explodes into two equal mass fragments. It is observed tha

t just after the explosion one fragment is traveling vertically up with speed v0 . What is the velocity of the other fragment? Hint: Velocity has both magnitude and direction.

1 answer:

yanalaym [24]3 years ago

3 0

Answer:

yeah

Explanation:

yeah yeah yeah yeah

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A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t

Pepsi [2]

Answer:

$W =84.6\ Nm$

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

$W = F\int_0^6 dx$

$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

hence, work done by the force is equal to $W =84.6\ Nm$

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A ball of a mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? Acceleration due

Vaselesa [24]

In order to solve this problem, there are two equations that you need to know to solve this problem and pretty much all of kinematics. The first is that d=0.5at^2 (d=vertical distance, a=acceleration due to gravity and t=time). The second is vf-vi=at (vf=final velocity, vi=initial velocity, a=acceleration due to gravity, t=time). So to find the time that the ball traveled, isolate the t-variable from the d=0.5at^2. Isolate the t and the equation now becomes $\sqrt{(2d)/a}$ . Solving the equation where d=8 and a=9.8 makes the time $\sqrt{(2*8)/9.8}$ =1.355 seconds. With the second equation, the vi=0 m/s, the vf is unknown, a=9.8 m/s^2 and t=1.355 sec. Substitute all these values into the equation vf-vi=at, this makes it vf-0=9.8(1.355). This means that the vf=13.28 m/s.

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3 years ago

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Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of LaTeX: 2.02.0

koban [17]

Answer:

$2.4\times 10^{-5} N$

Explanation:

We are given that

$I_1=I_2=2.0 A$

Distance between the wires, $d=5.0 mm=5\times 10^{-3} m$

1mm= $10^{-3} m$

Length of other wire,l=0.3 m

Length of one wire is infinite.

We have to find the magnitude of force between the two wires.

We know that magnetic force

$F=\frac{2\mu_0 I_1I_2 l}{4\pi d}$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

$F=\frac{2\times 10^{-7}\times 2\times 2\times 0.3}{5\times 10^{-3}}$

$F=2.4\times 10^{-5} N$

When the wires carrying equal current in opposite direction then the force between two wires repel to each other.

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3 years ago

Which of the following careers would physicists be least qualified to perform,

elena55 [62]

The answer to this question is D

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3 years ago

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38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago