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Lady bird [3.3K]
3 years ago
11

Water drips from a shower head (the sprayer at the top of the shower) and falls onto the floor 2.3 m below. The droplets are fal

ling at regular intervals (equal amounts of time between drops), with the first drop hitting the floor at the instant that the fourth drop starts to fall. When the first drop hits the floor (as the fourth drop is dripping from the shower head), how far below the shower head is the third drop? [The third drop is the one that fell right before the fourth drop.]

Physics
1 answer:
Maurinko [17]3 years ago
8 0

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.

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babunello [35]

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In case of air rushing out of balloon , the balloon experience a net force by the air coming out of it.

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3 0
3 years ago
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
A commuter train travels from New York to Washington, DC, and back in 6 hours and 5 minutes. The distance between the two statio
vovikov84 [41]

Answer:

The train's displacement is zero.

Explanation:

Given data,

The time taken by the train from NY to Washington and back is, t = 6 h 5 min

The distance between the two stations is, d = 363 km

Therefore, the total distance the train traveled is, d' = 726 km

The displacement is defined as the change in position coordinates with respect to its original position.

If the train travels from one point and returns back to the same point after some time, there is no change in the position coordinates with respect to its original position.

Hence, the train's displacement is zero.

3 0
4 years ago
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