All categories

3 years ago

Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6

Physics

1 answer:

inna [77]3 years ago

4 0

Answer:

Answer:

Q_1 = 7Q

Q_2 = 10Q

=10

Q_3 = 13.5Q

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

(9+1)

Q_1 = \frac{1}{4}*10Q

∗10

Q_1 = 2.5th\ itemQ

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

(7+7)

Q_1 = \frac{1}{2}*14Q

∗14

Q_1 = 7Q

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

(9+1)

Q_2 = \frac{1}{2}*10Q

∗10

Q_2 = 5th\ itemQ

=5th item

Q_2 = 10Q

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

(9+1)

Q_3 = \frac{3}{4}*10Q

∗10

Q_3 = 7.5th\ itemQ

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

(12+15)

Q_3 = \frac{1}{2}*27Q

∗27

Q_3 = 13.5Q

=13.5

You might be interested in

which of the following statements is correct when describing the forces at work on a bridge a. wires are in tension B. large pos

stiv31 [10]

Answer:

All the statements are true

Explanation:

A) Wires are in tension.

It is true because the loads exerted on the bridge are taken by each of the cables of the bridge. these cables transmit the force to the pillars or columns of the bridge. (check the image 01)

B) Large posts are in compression.

It is true each of the loads of the cables is tension, are transmitted to the columns of the bridge, in this way the balance will be maintained (check the image 01)

C) The bridge deck is both in tension and compression.

Yes it is true, the superior part of the deck will be in compression and the inferior part will be in tension, this can be seen in the image 02

8 0

3 years ago

A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total

NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$

6 0

3 years ago

Throw a paper ball towards a hard wall.

Len [333]

The answer is d hope this helps

7 0

3 years ago

23. The force required to accelerate a 2.0-kilogram

Marianna [84]

Answer:
C) 8.0 N

Explanation:
F=m*a

4 0

3 years ago

The function h = -t2 + 95 models the path of a ball thrown by a boy where h represents height, in feet, and t represents the tim

Evgesh-ka [11]

Answer:

A building where the boy is on the 12 floor

Explanation:

The equation of the height or position takes into account the initial conditions.

So the general equation is

x(t) = at+c

When you solve with the initial conditions t=0 you will find the position at the begining, so this is c the constant.

in our initial equation we have 95 as the constant. and this is a equation to find height so the initial height is 95feets.

Now a common building has a distance of 7.5 feet between every floor.

You divide 95 by 7.5 and you will find 12.6

6 0

4 years ago

Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6​

Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6