Assuming that this gas is in ideal state, we can use the relation that for every 1 mol of an ideal gas it would have a volume of 22.4 L. But before using this, relation we need to convert the number of grams of H2 into moles by using the molar mass of 2.02 g/mol.
moles H2 = 0.00922 g ( 1 mol / 2.02 g ) = 0.005 mol H2
Volume H2 at STP = 0.005 mol H2 ( 22.4 L / 1 mol ) = 0.102 L of H2
The acceleration is given as:
a = g sin(30°) where g is the gravitational acceleration
For g = 10 m/s^2, we get
a = 10 sin(30°) = 10 * 1/2 = 5 m/s^2
<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a.
Thus, d = [(0^2)-(15^2)]/(2*-7)
d = [0-(225)]/(-14)
d = 225/14
d = 16.0714 m
With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
<span>A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?</span>
