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tresset_1 [31]
2 years ago
10

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

to it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost?(Check 2 choices for the answer) *
Physics
1 answer:
erica [24]2 years ago
6 0

Explanation:

Given that,

Mass of a freight car, m_1=30,000-kg

Speed of a freight car, u_1=0.85\ m/s

Mass of a scrap metal, m_2=110,000\ kg

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J

Lost in kinetic energy is 8518.82. Negative sign shows loss.

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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

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3 years ago
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kondor19780726 [428]

Answer:

300 Pascal

Explanation:

Given

weight or force (F) = 6000 N

area (A) = 20 m²

pressure (p) = ?

we know

the force acting normally per unit area is pressure. So

P = F / A

= 6000 / 20

= 300 Pascal

Hope it will help :)

3 0
3 years ago
An airplanes lands with a velocity of +75 m/s and comes to a stop in 20 seconds. What is the acceleration of the airplane while
erma4kov [3.2K]

Answer:

The acceleration of the car, a = -3.75 m/s²

Explanation:

Given data,  

The initial velocity of the airplane, u = 75 m/s

The final velocity of the plane, v = 0 m/s

The time period of motion, t = 20 s

Using the I equations of motion

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                     a = (v - u) / t

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The negative sign indicates that the plane is decelerating

Hence, the acceleration of the car, a = -3.75 m/s²

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What are the statements?

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