Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa
Because the air inside the tires is kept at high pressure.
In fact, the force applied by the tires upwards to counter-balance the weight of the car (pushing downwards) is

where p is the pressure of the air inside the tires and A is the area of contact between the tire and the car. Therefore, a higher pressure means a larger force F, and eventually if the pressure p is higher enough the force F will be large enough to counterbalance the weight of the car.
Answer:

Explanation:
When the screw is turned by two turns then change in the length of the wire is given as




now we know by the formula of Young's modulus

so we have



Answer:
(A) = 3.57 m
Explanation:
from the question we are given the following:
diameter (d) = 3.2 m
mass (m) == 42 kg
angular speed (ω) = 4.27 rad/s
from the conservation of energy
mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1
where
Inertia (I) = 0.5mr^{2}
ω = \frac{v}{r}
equation 1 now becomes
mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}
gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}
4gh = 2v^{2} + v^{2}
h = 3v^{2} ÷ 4 g .... equation 2
from ω = \frac{v}{r}
v = ωr = 4.27 x (3.2 ÷ 2)
v = 6.8 m/s
now substituting the value of v into equation 2
h = 3v^{2} ÷ 4 g
h = 3 x (6.8)^{2} ÷ (4 x 9.8)
h = 3.57 m
Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.