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Rudik [331]
2 years ago
7

Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ign

ored) from their low-flying planes and scoring points for having the weights land close to a target. A plane 55m above the ground is flying directly toward a target at 48m/s .A)At what distance from the target should the pilot drop the weight?Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Arte-miy333 [17]2 years ago
4 0

Answer:

It should be dropped from 160.80m

Explanation:

To calculate the time it takes the weight to reach the ground from 55m we need to consider that the only acceleration applied to it is the gravity, the height at any time t can be calculate as:

Y=Y_{0}+Vy_{0}*t-1/2*g*t^{2}

If we consider that it starts from Y_{0}=55m to reach Y=0, and that the initial velocity on the vertical axis equals zero:

0=55m-1/2*g*t^{2}

solving for t:

t=\sqrt{\frac{55m*2}{g} } =\sqrt{\frac{110m}{9.8m/s^{2} } }=3.35s

Now to calculate the horizontal distance from the target:

X=X_{0}+Vx_{0}*t-1/2*a*t^{2}

We have to consider that:

It will travel from X_{0}=0m (we set the reference on the dropping point so the result X will be de needed distance;

As the air resistance can be ignored there is no force on the horizontal direction, a=0;

The starting velocity Vx_{0} is the same of the plane velocity

It will be traveling for a time t calculated before;

replacing this on the equation:

X=48m/s*3.35s=160.80m

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One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

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Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

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2 years ago
Which part of an atom makes up most of its volume ?
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Neutrons are also held together via the Strong Force, but don't have a charge so when separated, don't have an electromagnetic force pushing them away from each other.

However, electrons act differently. There is no "Strong Force" just the electromagnetic force. So, they keep a great distance from each other. 

So in an atom, protons and neutrons stay close to each other, taking up little volume, while electrons take up a lot of volume.

BTW, the reason why electrons and protons act differently when they are close together is because protons are made up of smaller particles the carry this Strong Force. For electrons, there is no smaller constituent. And therefore, all you have is the electromagnetic force to influence it. That's it.

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3 0
3 years ago
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

8 0
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