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neonofarm [45]
3 years ago
13

A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.

Physics
1 answer:
aleksley [76]3 years ago
6 0
The acceleration that the cheeseburger experienced is 20 m/s^2.
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What 2 parts of your foot do you use to dribble a soccer ball
Rasek [7]

Answer:

im not 100% sure but i think its the base of your big toe or the arch of your foot

Explanation:

thats how i do it

4 0
3 years ago
Gauss's law is usualy written as :
snow_tiger [21]

Answer:

(a) the net charge inside the closed surface.

Explanation:

In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.

4 0
3 years ago
In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occu
sasho [114]
Since there are four states, then the grand partition function of the system is

Z = 1+2 e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT}

where α is the chemical potential

Then, the occupancy of the system is

bar \ n=(e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT})/Z

Then using this equation, \alpha =-kT(V Z_{int} /N v_{Q}) and approximating Z_int to be kT/0.00018 eV, the model would look as that attached in the figure. That is the occupancy vs. pressure graph. 

There are more occupancies when the oxygen is high (high pressure) especially in the lungs. Heme sites tend to be occupied by oxygen. 

7 0
3 years ago
The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[
AysviL [449]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

P = \frac{E}{t}

where;

P is power in watts

E is energy in Joules

t is time in seconds

P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W

The average current that this cell phone draws when turned on:

P = IV

I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

5 0
3 years ago
Kind of mixture that has the same mixture throughout
Ainat [17]
It's a homogeneous mixture
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8 0
3 years ago
Read 2 more answers
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