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adelina 88 [10]
3 years ago
9

PLEASE HELP ON EDGE I WILL GIVE U A BRAINLIST IF U GET IT RIGHT

Physics
1 answer:
quester [9]3 years ago
6 0
It should be chestnut hair.

All boxes should be Ee which is dominant.
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The kinetic energy decreases 
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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.600 W/m2.
LenKa [72]

Answer:

gdgdrhhdhhsbsnjsjdgjdjdjsbdubjdbfn

Explanation:

nnndhdhjdjddjrjjdkdnfidjdbdjbdksjd

5 0
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The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert
andre [41]
2m X 2m = 4m ^2
960n / 4 = 240pa
3 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical
laila [671]

Answer:

I = 18 x 10⁻⁹ A = 18 nA

Explanation:

The current is defined as the flow of charge per unit time. Therefore,

I = q/t

where,

I = Average Current passing through nerve cell

q = Total flow of charges through nerve cell

t = time period of flow of charges

Here, in our case:

I = ?

q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C

t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s

Therefore,

I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)

<u>I = 18 x 10⁻⁹ A = 18 nA</u>

6 0
3 years ago
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