(The question and setup are below)

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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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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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]

The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]

If an object is lifted, work is done against gravitational force. The object gains energy.

Given are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.

The range of the projectile is given by R =  v²sin2θ / g

g = gravitational acceleration of Earth

If g = g(p) for planet , range  R =  v²sin2θ / g(p)..................(1)

The gravitational force of attraction = weight force

Gm² /d² = m g(p)

g(p) = Gm/d².........................(2)

For R = d/3, from equation (1), we have

d/3 =  v²sin2θ / g(p)

Plug the expression for g(p) , we get

v = √ [Gm/3dsin2θ ]

For velocity to be minimum, sin2θ =1

So, the minimum velocity will be

v = √ [Gm/3d]

Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]

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The speed of a 16cm wave with a period of 8 seconds is 2cm/s

Using the formula;

Speed = distance ÷ time

Distance = 16cm

time = 8 seconds

Substitute into the formula

Speed = 16 ÷ 8

Speed = 2 cm/s

Therefore, the speed of a 16cm wave with a period of 8 seconds is 2cm/s

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Answer:equal mass

Explanation:the mass of both is 5kg

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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The direction of this vector is 170° from the positive x-axis.

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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

Answer:

it is double the temperature change of iron

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

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

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⠀

[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]

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⠀

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

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⠀

The horizontal range of projectile at x.

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[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]

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⠀

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]

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⠀

[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]

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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

We can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.

The velocity time graph shows the change of velocity with time. Recall that velocity is the change of position with time.

As such, when we have a position time graph as shown, we can obtain the velocity from the graph and plot is against time to obtain a velocity time graph.

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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.

Whenever a wave is brought on by a cool summer breeze across the field. The grass tilts in the wind's direction before reverting to its original posture. Option A is correct.

Air pressure variations within our atmosphere result in the wind, which is the movement of air.

Areas with a low pressure are approached by air in high pressure. The air flows more quickly when there is a bigger pressure differential. Most rapid winds.

The wind is the movement of air produced by changes in air pressure in our atmosphere.

High-pressure air goes toward low-pressure locations. The bigger the pressure differential, the faster the air moves.

Hence, option A is correct.

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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 ]

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].

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.

Statement B  is true about the energy level of gas, Gases have more energy than liquids and solids.

A gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.

Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.

Gases have more energy than liquids and solids.

Hence statement B  is true about the energy level of gas.

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(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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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:

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

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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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 time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

The time taken for a motion to occur can be calculated using the following formula:

v² = u² - 2as

Where;

8.84² = 0² + 2 × a × 7.01

78.15 = 14.02a

a = 5.57m/s²

V = u + at

8.84 = 0 + 5.57t

t = 1.59s

Therefore, the time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.

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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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(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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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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