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zhannawk [14.2K]
3 years ago
7

How do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate?

Physics
1 answer:
Wittaler [7]3 years ago
3 0

Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force.  The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

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Alan rents a private jet for the weekend. He flies 400 km south to New York, then flies 700 km west to Chicago, then 1200 km to
Ratling [72]

Answer:

Total distance = 400+700+1200= 2300km

Explanation:

the resultant of d 1st right angle triangle + 1200

= 806.2 + 1200 = 2006.2km

7 0
3 years ago
A student runs 35 m east, then 12 m west. What is the distance run by the student?
lora16 [44]
47m total just add them up 35 + 12 = 47
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Two positive charges of 6 µC are separated by a distance of 50 cm in air. What is the electric field strength at the midpoint of
attashe74 [19]

Answer:

The electric field strength at the midpoint of the line joining the charges is zero (0)

Explanation:

Given that the two charges are both positive (same charge) and are equal in magnitude that is 6uC. The electric field strength at the midpoint of the line joining the two charges will be equal and opposite in magnitude, therefore they will cancel each other out and the electric field strength at this point will be equal to zero.

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3 years ago
A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
ArbitrLikvidat [17]

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
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Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0
3 years ago
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A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

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y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

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For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

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note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
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