Answer:
0.767m
Explanation:
We are given that the time interval between each droplet is equal.
We are also given that the fourth drop is just dripping from the shower when the first hits the floor.
If they fall at the same time interval and we know that the distance between the shower head and floor are the same, they must therefore fall at the same velocity.
The distance between each drop has to be the same given that they fall at equal time intervals.
Let this distance be x.
We can then partition the entire height of the system into three parts (as shown in the diagram).
Hence, we can say that:
x + x + x = 2.3m
3x = 2.3m
=> x = 2.3/3 = 0.767m
Therefore, at the time the first drop hits the floor, the third drop is only 0.767 m below the shower head.
Sound waves travel faster through <em>solids</em> than they do through gases or liquids. <em>(C) </em>They don't travel through vacuum at all.
Example:
Speed of sound in normal air . . . around 340 m/s
Speed of sound in water . . . around 1,480 m/s
Speed of sound in iron . . . around 5,120 m/s
Answer:
B) PbI2 + 2 KCl
Explanation:
To keep the the law of conservation of matter, the equation given above must be balanced i.e the total element in the reactant must be equal to the total elements in the product.
Given the equation
PbCl2 (aq) + 2 KI (aq) →
At the reactant shown, there are one mole of lead Pb, 2 moles of chlorine Cl, 2moles of Potassium K and 2 moles of Iodine.
During reaction, the Chlorine atom will react with the potassium atom K and the lead atom Pb will react with the iodine atom.
The resulting product that will balance the chemical equation is
PbI2 + 2 KCl
The equation will then become
PbCl2 (aq) + 2 KI (aq) → PbI2 + 2 KCl
If we look at both sides of the equation, we will see that all the elements have the same number of atoms.
Answer:
a) please find the attachment
(b) 3.65 m/s^2
c) 2.5 kg
d) 0.617 W
T<weight of the hanging block
Explanation:
a) please find the attachment
(b) Let +x be to the right and +y be upward.
The magnitude of acceleration is the same for the two blocks.
In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:
∑Fx=ma_x
T=m1a_x
14.7=4.10a_x
a_x= 3.65 m/s^2
c) <em>in order to calculate m we will apply newton second law on the hanging </em>
<em> block</em>
<em> </em>∑F=ma_y
T-W= -ma_y
T-mg= -ma_y
T=mg-ma_y
T=m(g-a_y)
a_x=a_y
14.7=m(9.8-3.65)
m = 2.5 kg
<em>the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax</em>
d) calculate the weight of the hanging block :
W=mg
W=2.5*9.8
=25 N
T=14.7/25
=0.617 W
T<weight of the hanging block
