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3 years ago

7

Why is there no drag in space?????? Help me please!!! Thanks!!!

2 answers:

dolphi86 [110]3 years ago

8 0

If there is no fluid, there is no drag. Drag is generated by the difference in velocity between the solid object and the fluid.

astraxan [27]3 years ago

7 0

It is beacuse of fluid If there is no fluid, there is no drag. Drag is generated by the difference in velocity between the solid object and the fluid. If this statement is correct then how can there be drag in space if there is no air?

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a 0.199 kg snowball moving west makes an inelastic collision with a 2.89 kg box moving 0.523 m/s west. afterward,they move west

kogti [31]

Answer:

The initial velocity of the snowball was 22.21 m/s

Explanation:

Since the collision is inelastic, only momentum is conserved. And since the snowball and the box move together after the collision, they have the same final velocity.

Let $m_1$ be the mass of the ball, and $v_1$ be its initial velocity; let $m_2$ be the mass of the box, and $v_2$ be its velocity; let $v_f$ be the final velocity after the collision, then according to the law of conservation of momentum:

$m_1v_1+m_2v_2=v_f(m_1+m_2)$ .

From this we solve for $v_1$ , the initial velocity of the snowball:

$\boxed{v_1=\frac{v_f(m_1+m_2)-m_2v_2}{m_1}}$

now we plug in the numerical values $m_1=0.199\:kg$ , $m_2=2.89\:kg$ , $v_2=0.523\:m/s$ , and $v_f=1.92\:m/s$ to get:

$v_1=\frac{1.92*(0.199+2.89)-2.89*0.523}{0.199}$

$\boxed{v_1=22.21\:m/s}$

The initial velocity of the snowball is 22.21 m/s.

<em>P.S: we did not take vectors into account because everything is moving in one direction—towards the west.</em>

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Alexus [3.1K]

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3 years ago

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BlackZzzverrR [31]

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2 years ago

A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due

Korvikt [17]

Answer:

$\Phi_E=0.11\frac{N\cdot m^2}{C}$

Explanation:

According to Gauss's Law, the electric flux of a charged sphere is the electric field multiplied by the area of the spherical surface:

$\Phi_E=EA\\\Phi_E=E(4\pi R^2)\\\Phi_E=\frac{q}{4\pi \epsilon_0 R^2}(4\pi R^2)\\\Phi_E=\frac{q}{\epsilon_0}$

This is identical to the electric flux of a point charge located in the center of the sphere.

$\Phi_E=\frac{1*10^{-12}C}{8.85*10^{-12} \frac{C^2}{N\cdot m^2}}\\\Phi_E=0.11\frac{N\cdot m^2}{C}$

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3 years ago