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Gre4nikov [31]
3 years ago
15

In a very busy off-campus eatery one chef sends a 235-g broccoli-tomato-pickle-onion-mushroom pizza sliding down the counter fro

m left to right at 1.65 m/s. Almost simultaneously, the other chef launches a 351-g veggieburger-on-a-bun (with everything, but hold the horseradish) along the same counter from right to left at 2.47 m/s. The two delicacies collide head on at the given speeds–the counter is practically friction-free due to accumulated grease–and merge into a single, unbelievably savory serving. In what direction does the delectible dish move, if it moves at all?
Physics
1 answer:
STALIN [3.7K]3 years ago
3 0

Answer:

0.47922 kgm/s and will move right to left

Explanation:

m_1 = Mass of broccoli-tomato-pickle-onion-mushroom pizza = 235 g

m_2 = Mass of veggieburger-on-a-bun = 351 g

v_1 = Velocity of broccoli-tomato-pickle-onion-mushroom pizza = 1.65 m/s

v_2 = Velocty of veggieburger-on-a-bun = -2.47 m/s

Left to right is considered positive direction and the opposite is negative direction

Net momentum is

p=m_1v_1+m_2v_2\\\Rightarrow p=0.235\times 1.65+0.351\times -2.47\\\Rightarrow p=-0.47922\ kgm/s

The combined dish has a momentum of 0.47922 kgm/s and will move right to left

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$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

    = 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

                                           $ =\frac{1}{2} \times 1.225  \times 200^2$

                                          $=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

     $=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

     = 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

                  $=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

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7 0
2 years ago
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

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I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
2 years ago
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