Answer:
You could throw away the following items:
Mixed paper: Newspaper, magazine, junk mail, cardboard, frozen food packaging, wrapping paper, paper bags, etc.
Cans: aluminum, foil, tin, steal, aerosol (empty without pressure) and metal food trays. ...
Unbroken glass: Wine bottles, bottles and jars.
Plastic: Rigid containers like milk jugs, shampoo or water bottles.
Hope this helped
We know, the ideal gas equation,
P1V1 / T1 = P2V2 / T2
Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K
P2 = 400 mm Hg
T2 = -23 + 273 = 250 K
Substitute their values,
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83
In short, Your Answer would be approx. 15.83 m3
Hope this helps!
The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this
Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²
Height = (4.9 m/s²) (5.76 s²)
Height = (4.9/5.76) meters
Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)
Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2

Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;

T = 7.83 X10⁻⁷ s