Answer:
1.29 kg
Explanation:
As we know that
W= FS
1.56 = F (0.78)
F= 1.56/0.78
F= 2 N
Now, According to the definition of Force
F= ma
then
m= F /a
m= 2 / 1.54
m = 1.29 kg
Answer:
0.10013 atm
Explanation:
Applying Boyle's Law,
P'V' = PV................... Equation 1
Where P' = Initial pressure of air, V' = Initial volume of air, P = Final pressure of air, V = Final volume of air.
make P the subject of the equation
P = P'V'/V..................... Equation 2
Given: P' = 0.355 atm, V' 0.110 m³, V = 0.390 m³
Substitute into equation 2
P = 0.355(0.11)/0.39
P = 0.10013 atm.
Answer:
In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.
Explanation:
The current flowing through your system is described by Ohm's law
V = I R
where I is the current, V the voltage and R the resistance
in this case three barateras are taken in series giving a total voltage of V = 4.5 V the typical resistance values of dry skin is R = 1000 000Ohm and the resinification of wet skin is R = 100000 ohm
let's calculate the current flowing
I = V / R
I = 4.5 / 1000000
I = 4.5 10⁻⁶ A
this is the current with dry hands, we see that much less than the value that allows to feel a painful response by the body
If the skin is
I = 4,5 / 100,000
I = 4.5 10⁻⁵ A
This value is small, but it is close to the pain threshold, but it is in the range of slight discomfort.
In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.
Answer:
There are basically seven spectral types which are used for classifying the stars and it is arranged in the order of increasing temperature from the hottest temperature to the coldest division are: O, B, A, F, G, K, and M. These categories are basically divided into the sub category in the spectral where O is the hottest type and M is the coldest type.
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
(2)
where
is the refractive index of the second medium and
is the refractive index of the first medium.
We can find the ratio
by using Snell's law:
(1)
where
is the angle of incidence
is the angle of refraction
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
