Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,

The expression of
will be,

Now put all the values of partial pressure, we get


Therefore, the partial pressure of
is, 67.009 atm








<h3>☯ <u>By using formula of Lens</u> </h3>











<h3>☯ <u>Now, Finding the magnification </u></h3>





<h3>☯ <u>Hence</u>,

</h3>


Answer:
E = 307667 N/C
Explanation:
Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.
This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:
0.0098 N = T * cos(37)
then T = 0.0098/cos(37) N = 0.01227 N
Knowing the tension's magnitude, we can find its horizontal component:
T * sin(37) = 0.007384 N
and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:
0.007384 N = E * 2.4 * 10^(-8) C
Then E = 0.007384/2.4 * 10^(-8) N/C
E = 307667 N/C
This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)
Answer:
Final velocity = 7.677 m/s
KE before crash = 202300 J
KE after crash = 182,702.62 J
Explanation:
We are given;
m1 = 1400 kg
m2 = 4700 kg
u1 = 17 m/s
u2 = 0 m/s
Using formula for inelastic collision, we have;
m1•u1 + m2•u2 = (m1 + m2)v
Where v is final velocity after collision.
Plugging in the relevant values;
(1400 × 17) + (4700 × 0) = (1400 + 1700)v
23800 = 3100v
v = 23800/3100
v = 7.677 m/s
Kinetic energy before crash = ½ × 1400 × 17² = 202300 J
Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J