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2 years ago

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An 80 kg skydiver is falling at terminal velocity. What is the value of air resistance acting on his body? Consider, what are th

e knowns and unknowns?

1 answer:

Mrac [35]2 years ago

5 0

Answer:

800 N

Explanation:

During terminal velocity, air resistance=weight

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A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo

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Answer:

$\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$

Explanation:

The total force on the particle is given by

$\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}$

Then, by replacing we have:

$q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N$

where the cross product can be made with the determinant method.

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3 years ago

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Why are yet planes faster tharn. cargo planes?

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Answer:

cargo planes hold cargo so there hevier

Explanation:

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2 years ago

A massless string connects a 10.00 kg mass to a 13.00 kg cart which is resting on a frictionless horizontal surface. The mass ha

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The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

<h3>Acceleration of the cart</h3>

The acceleration of the cart is determined from the net force acting on the mass-cart system.

Upward force = Downward force

ma = mg

13a = 10(9.8)

13a = 98

a = 98/13

a = 7.54 m/s²

Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².

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The correct answer is B :)

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3 years ago

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A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

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2 years ago