C1=10 μF , C2= 50 μF, C3=9 μF, C4=2 μF, C5= 0.1 μF

What is the equivalent capacitance of these capacitors as they are arranged?

The height of the image is -6 cm.

The reflecting surface of a concave mirror is curled inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given that, an object placed 36 cm from a concave mirror having the focal length of 12 cm and height of image is 2.50 cm

We have to find height of the Image

Using Mirror Formula:

1/f = 1/v + 1/u

Where,

f = focal length

v =  Image distance from the mirror

u = object distance from the mirror (concave)

Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.

1/(-12) = 1/v + 1/(-36)

1/(-12) = 1/v - (1/36)

1/v = -0.055

v = -18 cm

Therefore, Image distance the from the concave mirror is -18 cm.

Now,

m = -v/u = hi/h

Substitute the values,

-(-18)/(-36) = hi/12

-0.5 = hi/12

-6 = hi

Therefore, the height of the image is -6 cm.

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Answer:equal mass

Explanation:the mass of both is 5kg

A valve placed is before the rotameter to regulate the flow of liquid entering the extraction bed. The correct option is c.

Control valves are generally used to maintain a flow as close as possible to the desired value.

Valves positioned before the rotameter for throttling purposes. A control valve will regulates the rate of fluid flow. Control valves control typically flow rate, pressure, and temperature.

Thus, the correct option is c.

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Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

Answer:

Fritional force

Explanation:

Fritional force is a force that always opposes the motion of an object and is frequently ignored

C

Explanation:

Applied force, air resistance

Typing in the commands and instructions and pressing enter will put a new command in the command prompt​.

This refers to a words or phrase which causes the computer to execute certain tasks or functions.

Typing the exact instruction and entering it will ensure that the chosen tasks are found in the command prompt.

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The time taken for the sailfish to reach the herring if it is 55.0 m away is 2.7 s.

The time taken for the star fish to travel the given distance is calculated as follows;

s = ut + ¹/₂at²

55 = 11.8t + (0.5)(6.39)t²

55 = 11.8t + 3.195t²

3.195t² + 11.8t - 55 = 0

solve the quadratic equation using formula method,

where;

a = 3.195, b = 11.8, c = -55

t = 2.7 s

Thus, the time taken for the sailfish to reach the herring if it is 55.0 m away is 2.7 s.

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Answer:

Multiple so you can test multiple hypothesis at once

Explanation:

because

Answer:

To test the hypothesis, we need to make an observation or perform an experiment associated with the prediction. For instance, in this case, we would plug the toaster into a different outlet and see if it toasts.

inspired by answer up there

sorry i couldn't add details because think its bad

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

The resultant of the two forces is about 170 N.

The term resultant force has to do with the single force that has the same magnitude and direction as two or more forces acting together.

In this case, the both forces are acting in the forward direction. This implies that the resultant force is 90 N  + 80 N = 170 N.

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The force result in stretching the spring 10.0 centimeters is 2.5N.

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m  / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

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Answer: 2.5 newtons

source- trust me bro

(a) From the first sentence, particularly goes with condition.

(b) From the second sentence, everybody goes with is.

(c) From the third sentence, forecasts goes with enable.

In subject- verb agreement, if a subject is singular, its verb must also be singular and if a subject is plural, its verb must also be plural.

From the first sentence, the correct subject-verb agreement is;

The release of energised particles from the sun , coupled with particularly helpful overnight condition, are predicted to make the "Northern Lights" visible much further south than normal.

From the second sentence, the correct subject-verb agreement is;

'Everybody is talking about it. It's a once-in-a-decade opportunity that people really shouldn't miss.'

From the third sentence, the correct subject-verb agreement is;

'The forecasts given out by the weather office enable us to know when the skies will be clear.

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Answer:

the answer is ambivert have a good fourth of July

Answer:

i believe the answer is ambivert

It is defined as the stress that results in the elongation of the material

To deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.

The power generated in a circuit is directly proportional to the current flowing in the circuit.

P = IV

where;

Voltage in the circuit also increases with increase in the current flowing in the circuit.

Thus, to deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.

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Answer:

The rod is stretched because there is a larger gravitational force on the mass closest to mass M.

F = G M m / r^2 - G M m / (r + l)^2       the difference of the forces on m's

F = G M m [1 / r^2 - 1 / (r + l)^2]

[1 / r^2 - 1 / (r + l)^2] = (r^2 + 2 r l + l^2 - r^2 / [r^2 (r + l)^2]

= (2 r l + l^2) / [r^2 (r + l)^2]

if l is small compared to r = (2 r l ) / [r^2 (r + l)^2]

= (2 r l ) / [r^4 + 2 r^3 l ]

The number of significant digits to the answer of the following problem is four.

The number of digits rounded to the approximate integer values are called the significant digits.

The following problem is

(2.49303 g) * (2.59 g) / (7.492 g) =

On solving we get

= 0.86184566204

The answer is approximated to  0.86185

Thus, the significant digits must be four.

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[tex]\textsf {C. Force}[/tex]

[tex]\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}[/tex]

(a) The time for a particle of water leaving the orifice to reach point B is 0.63 s.

(b) The horizontal distance x where it hits from surface is 3.94 m.

The time of the water to reach the surface is calculated as follows;

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

where;

t = √( (2 x 2)/(9.8) )

t = 0.63 s

The horizontal distance is calculated as follows;

X = vt

where;

X = (√2gh)t

X = (√2 x 9.8 x 2) x 0.63

X = 3.94 m

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Answer:

it is double the temperature change of iron

The amount of time it will take to have 125 grams of erbium-165 left

in the sample is; C: 31.2 hours

The formula for amount of substance after decay is;

N(t) = N₀(¹/₂)^(t/t_¹/₂)

We are given;

Half life; t_¹/₂ = 10.4 hours

Initial amount; N₀ = 1000 g

Amount left; N(t) = 125 g

Thus;

125 = 1000 * (¹/₂)^(t/10.4)

125/1000 = (¹/₂)^(t/10.4)

In 0.125 = (t/10.4) In 0.5

-2.08 = -0.693(t/10.4)

t/10.4 = -2.08/-0.693

t = 10.4 * 3

t = 31.2 hours

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Answer:

from my side the will be b only the mass of the planet thank you

Answer:

D

Explanation:

W=mg

g=GM/r²(G=Gravitational constant, M=mass of earth, r=Radius of earth)

>>W=m(GM/r²)

W is directly proportional to mass of earth and mass of your body

Answer

688.32m and 277.44m

Explanation :

⠀

[tex] \large{\maltese{\textsf{\underline{To find :-}}}} [/tex]

The X and Y coordinates of the rocket relative of firing

⠀

⠀

[tex] \large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s [/tex]

⠀

⠀

[tex] \Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}} [/tex]

⠀

⠀

The horizontal range of projectile at x.

⠀

[tex] \sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t} [/tex]

⠀

⠀

[tex] \large\textsf{\underline{Now substituting the required values}} [/tex]

⠀

⠀

[tex] \sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}} [/tex]

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

[tex]\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2} [/tex]

⠀

⠀

[tex]\textsf{ \large {\underline{Now substituting the required values}} } [/tex]

⠀

⠀

[tex] \sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}} [/tex]

⠀

⠀

⠀

Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

Answer:

Approximately [tex]14\; {\rm N}[/tex].

Explanation:

By Coulomb's Law, the magnitude of the electrostatic force between two charges is proportional to the product of the magnitudes of the two charges.

For example, consider charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex] that are apart from one another by a distance of [tex]r[/tex] in between. Let [tex]k[/tex] denote Coulomb's constant. By Coulomb's Law, the magnitude of electrostatic force between the two charges would be:

[tex]\displaystyle F = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex].

In this question, the product of the magnitude of the two charges was originally [tex]3\; {\rm \mu C} \times 7\; {\rm \mu C} = 21\; {\rm (\mu C)^{2}}[/tex]. After [tex](-1\; {\rm \mu C})[/tex] is added to each charge, product of the magnitude of the two charges would become [tex](3 - 1)\; {\rm \mu C} \times (7 - 1)\; {\rm \mu C} = 12\; {\rm (\mu C)^{2}}[/tex].

Thus, the product of the magnitude of the two charges has been scaled to [tex]12\; {\rm \mu C}[/tex] from [tex]21\; {\rm \mu C}[/tex] . The magnitude of the electrostatic force between the two charges would be scaled from [tex]25\; {\rm N}[/tex] to [tex]25\; {\rm N} \times (12 / 21) \approx 14\; {\rm N}[/tex].

At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².

The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.

The concentration of H⁺ ions is calculated as:

pH = -log [H⁺]

2 = -log [H⁺]

[H⁺] = 10⁻²

The concentration of OH⁻ ions is calculated as:

pH + pOH = 14

pOH = 14 - 2

pOH = 12

Solving further:

pOH = -log [OH⁻]

12 =-log [OH⁻]

[OH⁻] = 10⁻¹²

At a pH = 2, the ratio of hydrogen ions to hydroxide ions:

10⁻² ÷ 10⁻¹² = 10¹⁰

Therefore, at pH = 2 the ratio is 10¹⁰.

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The flux of [tex]\vec E = -x\,\vec\imath + y\,\vec\jmath[/tex] is given by the surface integral

[tex]\displaystyle \iint_S \vec E \cdot d\vec\sigma[/tex]

where [tex]S[/tex] is the given square region, which we can parameterize by

[tex]\vec s(x, z) = x\,\vec\imath + z\,\vec k[/tex]

with [tex]0\le x\le 1[/tex] and [tex]0\le z\le 1[/tex]. The area element is

[tex]d\vec\sigma = \vec n \, dx\,dz[/tex]

where [tex]\vec n[/tex] is the normal vector to [tex]S[/tex]. Depending on the orientation of [tex]S[/tex], this vector could be

[tex]\vec n = \dfrac{\partial\vec s}{\partial x} \times \dfrac{\partial\vec s}{\partial z} = -\vec\jmath[/tex]

or [tex]-\vec n = \vec \jmath[/tex]; either way, the integral reduces to

[tex]\displaystyle \iint_S \vec E \cdot d\,\vec\sigma = \int_0^1 \int_0^1 (-x\,\vec\imath + z\,\vec k) \cdot (\pm\vec\jmath) \, dx\,dz = \boxed{0}[/tex]

Explanation:

it is almost zero .this is because the distance and the electrostatic force are inversely proportional

a)The velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec

b) The time taken for the object to reach the ground will be 11.4 × 10³ sec.

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The velocity in the x-direction;

[tex]\rm v_x = (vcos \theta)^2 +2gh \\\\ v_x = (50 cos 30)^2+ 2 \times 9.81 \times 4000 \\\\ v_x = 80355 m/sec[/tex]

Velocity in y-direction;

[tex]\rm v_ y = (vsin \theta)^2 + 2gh \\\\ v_ y = (25)^2+2 \times 9.81 \times 4000 \\\\ v_y =79105 \ m/sec[/tex]

The resultant velocity is found as 1.12 × 10⁵ m/sec.

The time taken to reach the ground is found as;

[tex]\rm v = u+gt \\\\ 1.12 \times 10^5 \ m/sec = 50 \ m/sec + 9.81 \times t \\\\ t = 11.4 \times 10^3 \ sec[/tex]

Hence, the velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec, and the time taken for the object to reach the ground will be 11.4 × 10³ sec.

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The work done in pulling the sled is 445,930.2 Joules.

The acceleration in the x direction, assuming that friction is negligible, is  20.85 m/s².

Time taken to pull the sled 185 m is 4.21s.

Work done is equal to product of force applied and distance moved.

Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.

Work = Force x Distance x cos(angle)

W= 1225 x 385 x cos 19°

W = 445,930.2 Joules

Thus, the  work done in pulling the sled is 445,930.2 Joules

From the Newton's second law of motion, we have'

F = ma

acceleration, a = 1225cos19° / ( 545 /9.81)

a = 20.85 m/s²

Thus, the acceleration in the x direction is  20.85 m/s²

Using the second equation of motion, we get

s = ut+ 1/2 at²

Substitute the values, we have

185 m = 0 +1/2 x 20.85 x t²

t = 4.21 s

Thus, the time taken to pull the sled 185 is 4.21s.

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The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

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