Which material BEST allows electricity to pass through it?

A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr

Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration $a=g=9.8 m/s^2$ directed downward, initial vertical position $d=750 m$ , and initial vertical velocity $v_0 = 0$ . We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

$v_f^2 -v_i^2 =2ad$

substituting, we find

$v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s$

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

$v_f=\frac{121.2 m/s}{2}=60.6 m/s$

In order to do that, we use again the same SUVAT equation substituting $v_f$ with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

$v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m$

Which means that the heigth of the packet was

$h=750 m-187.4 m=562.6 m$

3 0

2 years ago

Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of

Andre45 [30]

Well first of all, I think the students may have been correct.
If they didn't use distilled water, and if it wasn't exactly at
standard temperature, then the mass of 25.0 mL could
very well be 25.4 grams. We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

+0.4 / 25.0 = +0.016 .

To change a decimal to a percent, move the
decimal point two places that way ===> .

+ 0.016 = +1.6 % .

Their measurement was 1.6% too high.

Let's not call it an 'error'. Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value. OK ?

4 0

3 years ago

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A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$

3 0

3 years ago

two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change

Verdich [7]

The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together

5 0

3 years ago

For questions 1-10 fill in the blank with the letter of the description that best matches the term

tigry1 [53]

B-Pitcher C-Catcher H-Strike I-Umpire G-Strike Zone E- Foul Ball F- Ball J- Pick-off D-Error A- Shortstop. I think (Sorry for them being out of order. I had to break them down)

4 0

3 years ago

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