Tenemos.
Masa de la mesa = 20kg
Masa encima = 10kg
Masa total = 30kg
Area de cada pata = 20cm² = 20cm² * 1/10000cm² * 1m² =0,002m²
Presió(P)n que se ejerce sobre cada pata.
P =F/A
P = Masa por gravedad/A Masa = 30kg Gravedad =9,8m/s²
P =(30kg * 9,8m/s²)/0,002m²
P = 294kg * m/s²/0,002m² Pero kg *m/s² = Nw
P = 294Nw0,002m²
P = 147000 Nw/m² Pero Nw/m² = pa
p =1,47 * 10⁵ pa
Respuesta.
La presión que se ejerce sobre cada pata es de 1,47 * 10⁵pa
Two half lives so it is 4000 years old
Answer: B
Explanation:
Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m.
The parameters given from the question are:
Mass M = 2kg
Height h = 20m
Let g = 9.8m/s^2
At the bottom of the incline plane, the object will experience maximum kinetic energy.
From conservative of energy, maximum K.K.E = maximum P.E
Maximum P.E = mgh
Maximum P.E = 2 × 9.8 × 20 = 392 J
But
K.E = 1/2mv^2
Substitute the values of energy and mass into the formula
392 = 1/2 × 2 × V^2
V^2 = 392
V = sqrt( 392 )
V = 19.8 m/s
V = 20 m/s approximately
1. Velocity
2. Time
3. Idk
4. Idk
5. D. I think it may be A. but I think D.
To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
