A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 70 Ω, R2 = R6 = 106 Ω R3 = 59 Ω, and R4 = 83 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is I3?

Answer:

Let distance between two cities is x km . So total distance travelled bus x+x=2x km and time taken was=(x/40)+(x/60)=(3x+2x)/120=5x/120=x/24 hrs . So avg speed is 2x/(x/24)=48 km per hr.

or

dt1=60km1h

dt2=30km1h

2dt1+t2=r¯

d=60t1

d=30t2

60t1=30t2

2t1=t2

2dt1+(2t1)=r¯

2(60t1)t1+2t1=r¯

2(60t1)=r¯(t1+2t1)

120=r¯+2r¯

r¯=1203

r¯=40km/h

The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.

Three equation of motion are:-

Where v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.

In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.

Applying equation 1 to find the initial speed of plane

v = u + at

0 = u + (-6.34 × 5400)   {v=0 as plane will stop after 5400 sec}

u =  6.34 × 5400

u = 34236 m/s

Initial velocity of plane is 34236 m/s

Applying equation 2 to find the displacement of plane in that time period

s = ut + (1/2)at²

s = ( 34236 × 5400 )  - ( (1/2) × 6.34 × 5400² )

s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )

s = 5400 × ( 34236 - 17118 )

s = 5400 × 17118 metres

s = 5.4 × 17118 Km

s = 92437.2 Km

Distance travelled by plane is 92437.2 Km

So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.

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Answer:

500 g

Explanation:

Specific heat of water = 1 j/g-c

Heat given up by hot water = 200(75-25)(1) = 10 000 j

this is the heat GAINED by the cold water

10000 j =  x ( 25 -5)(1)  

          x =500g

Explanation:

hi

please help me. ineed answers

Gerund/Infinitive

1. Put the verbs in brackets into the correct infinitive form or the -ing form.

1. Jane went on____(sleep) for another two hours.

2. He told us his name and went on____ (introduce) us to his wife.

3. She tried (finish) her homework, but it was too difficult.

4. You should try (eat) more fruit. It's good for your health.

5. He regrets (argue) with his best friend.

6.We regret (inform) you that tonight's performance will be cancelled.

7. Oh, no! I forgot____(lock) the front door.

8. I'll never forget_(meet) my favourite film star.

9.Claire likes (ski). She says it's very exciting.

10. I like__(go) to the dentist every six months.

11.I must remember (post) these letters today.

12. I remember__(read) the book, but I don't know who wrote it.

13. I'm sorry for___(forget) your birthday. It was awful of me.

14. I'm sorry.(say) that you have failed the exam.

15. She is afraid___(climb) the tree in case she falls.

16.Mary never wears her diamond ring. She is afraid of (lose) it.

17. I have stopped . (buy) some food before continuing our journey.

18. We stopped (watch) horror films because they give me nightmares

19. Tom stopped__(pick up) his washing on the way home.

20 If you don't stop. (smoke), you'll make yourself ill.

21.Will you quit___(complain)! It doesn't help.___(solve) the problem.

22.If you ever decide___(sell) your car, let me____(know).

23.I would like you.__(water) the plants for me at the weekend.

24.I clearly remember__(set) my alarm clock before__(go) to bed last night.

25. These plants require(water) every day.

26. I resent you (speak) to me like that! Have some respect.

27. It would be good for the children__ (play) outdoors more often.

28. I promised __(take) Jill to the party, but I don't feel like__(go) now.

29. Don't waste your time (look for) the document.Ask Mr Gale.

30. Please, excuse his__(leave) so early. He wants __(catch up) with his (study).

31. Try__(phone) John at the office if he's not at home.

32. I tried my best__(finish), but there just wasn't enough time.

33. He was promoted in 1990 and went on__(become) a company director.

34. The band went on__(play) even after the lights had gone out.

35. "Why is the baby crying?" "I think he wants (feed)."

36. Sharon wants__(talk) to you.

37. Jane was afraid__(show) her school report to her parents.

38. I'm afraid of__(lose) my way in the forest.

39. I regret__ (inform) you that your husband has been arrested.

The crest is the highest point of a pebble-induced ripple in a body of water. Option a is correct.

The distance between two successive troughs or crests is known as the wavelength. The highest point of the wave is the crest, while the trough is the lowest.

The high point of the ripple seen in a pond after throwing in a pebble is the crest.

Hence option a is correct.

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The cost to use all lamps is $62.

           power=energy/time

          (60+ 100+150)=energy/2.5

          energy=775 joule

cost for 1 hour=$0.08

                      = 775 * 0.08

                     = $62

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Answer:

6.2 cents

Explanation:

(60+100+150/1000)*2.5*8=6.2 cents

Answer:

0.4

Explanation:

ω = km

ω = 1000 x 4

ω = 4000 divide this by 10000 and you get 0.4s

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 radius of a helium atom 4.8 times bigger  then the radius of an alpha particle.

The structure of alpha particles, also known as alpha rays and alpha radiation, is similar to that of the helium-4 nucleus and has been made up of two protons as well as two neutrons bonded together.

For welding metals like aluminum, helium was utilized as an inert gas environment. It is also employed in rocket propulsion.

The radius of a helium atom 4.8 times bigger  then the radius of an alpha particle. More precisely than it's ever been, the diameter of the helium atom's nucleus, the alpha particle, had also been measured. Outcomes just point to a size of 1.6782 femtometers, 4.8 times more accurate than earlier readings.

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Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by

[tex]x = v_i \cos(\theta_i) t[/tex]

[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]

and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are

[tex]v_x = v_i \cos(\theta_i)[/tex]

[tex]v_y = v_i \sin(\theta_i) - gt[/tex]

The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when

[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]

which means

[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]

At the same time, the ball will have traveled half its horizontal range, so

[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]

Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :

[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]

Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)

[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]

Now,

[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]

so it follows that (d)

[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]

Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is

[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]

Finally, in the work leading up to part (e), we showed the time to maximum height is

[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]

but this is just half the total time the ball spends in the air. The total airtime is then

[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]

and the ball is in the air over the interval (a)

[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]

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 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]

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.

Answer:

Approximately [tex]53^{\circ}[/tex], assuming that the upper and lower surfaces of the glass on this boat are parallel.

Explanation:

Assume that the upper and lower surfaces of the glass at the bottom of this ship are parallel. Refer to the diagram attached. The two normals would also be parallel to each other.

The following angles would be alternate interior angles between the two normals:

Since the two normals are parallel to each other, these two angles would have the same value. Let [tex]\theta_{\text{glass}}[/tex] denote the value of both of these angles.

Let [tex]\theta_{\text{src}}[/tex] denotes the angle at which a beam of light leaves the original medium (angle of incidence.) Let [tex]\theta_{\text{dst}}[/tex] denote the angle at which this beam of light enters the new medium.

Let [tex]n_\text{src}[/tex] and [tex]n_\text{dst}[/tex] denote the refractive indices of the original and the new medium, respectively. By Snell's Law:

[tex]\begin{aligned}\frac{\sin(\theta_{\text{dst}})}{\sin(\theta_{\text{src}})} = \frac{n_{\text{src}}}{n_{\text{dst}}}\end{aligned}[/tex].

Let [tex]\theta_{\text{water}}[/tex] denote the angle at which the beam of light in this question leaves the water. Let [tex]\theta_{\text{air}}[/tex] denote the angle at which this beam of light enters the air. It is given that [tex]\theta_{\text{water}} = 37^{\circ}[/tex], while [tex]\theta_{\text{air}}[/tex] is the value that needs to be found.

Let [tex]n_{\text{air}}[/tex], [tex]n_{\text{water}}[/tex], and [tex]n_{\text{glass}}[/tex] denote the refractive index of air, water, and glass, respectively. By Snell's Law:

[tex]\begin{aligned}\frac{\sin(\theta_{\text{glass}})}{\sin(\theta_{\text{water}})} = \frac{n_{\text{water}}}{n_{\text{glass}}}\end{aligned}[/tex].

[tex]\begin{aligned}\frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{glass}})} = \frac{n_{\text{glass}}}{n_{\text{air}}}\end{aligned}[/tex].

Thus:

[tex]\begin{aligned} & \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{water}})} \\ =\; & \frac{\sin(\theta_{\text{glass}})}{\sin(\theta_{\text{water}})}\times \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{glass}})} \\ =\; & \frac{n_{\text{water}}}{n_{\text{glass}}}\times \frac{n_{\text{glass}}}{n_{\text{air}}} \\ =\; & \frac{n_{\text{water}}}{n_{\text{air}}}\end{aligned}[/tex].

Since [tex]\theta_{\text{water}} = 37^{\circ}[/tex]:

[tex]\begin{aligned} & \sin(\theta_{\text{air}})\\ =\; & \sin(\theta_{\text{water}}) \times \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{water}})} \\ =\; & \sin(\theta_{\text{water}})\times \frac{n_{\text{water}}}{n_{\text{air}}} \\ =\; & \sin(37^{\circ}) \times \frac{1.33}{1.0} \\ \approx \; & 0.800 \end{aligned}[/tex].

Therefore:

[tex]\begin{aligned}\theta_{\text{air}} &= \arcsin(\sin(\theta_{\text{air}})) \\ & \approx \arcsin(0.800) \\ &\approx 53^{\circ} \end{aligned}[/tex].

In other words, this beam of light would leave the glass at approximately [tex]53^{\cic}[/tex] from the normal.

Answer:

1 / f = 1 / o + 1 / i = (i + o) / o * 1

f = o * i / (o + i) = 60 * 30 / (60 + 30) = 1800 / 90 = 20 cm

Both the object and image are in positive space for a mirror

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.

When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.

When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.

So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

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Answer:

Following precautions for obtaining the refractive index of a triangular glass prism:-

Explanation:

(1) All the faces of the prism should be neat and clean.

(2) Pins for holding the paper on the drawing board must be pinned perpendicular to the paper for better handling.

(3) While fixing the pins for checking the refractive index of the prism, make sure that the reflective images of the pins should be aligned to your eye to avoid any type of parallax error.

(4) Pins should not be removed or pinned again during the experiment.

(5) Avoid mishandling of the prism.

(6) Same edge of the prism should be taken as vertex for observations.

(7) For drawing the boundary of the prism, a sharp pencil should be used.

(8) Soft board and pointed pins should be used.

(9) The distance between the pins should be 5 cm or more.

(10) Make sure the glass slab or prism taken are polished and not broken.

Hope this helps!

Please mark me as brainlist.

Based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.

From the standpoint of  brain lateralization, differences do exist such as based on experience such as handedness or specific skills such as playing a guitar.

Note that Left-handers are said to have reduced or little lateralized brains, which tells us that the two halves of the brain are little different than as seen in the right-handers.

Therefore, I can say that based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.

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Answer:

Tolerate

Explanation:

Answer:

Joes displacement is 54m

The voltage of the secondary will be 36 V.From the given conditions primary transformer has 3 times as many turns in the secondary coil.

Electromagnetic induction is what causes the induced voltage. Electromagnetic induction is the process of generating emf (induced voltage) by subjecting a conductor to a magnetic field.

In this case, a magnet is pushed in and out of a wire coil attached to a high-resistance voltmeter.

Typically, a transformer's primary winding is attached to the input voltage source and changes electrical power into a magnetic field.

The secondary winding's role is to turn this alternating magnetic field into electricity, generating the necessary output voltage.

Given data;

Primary coil voltage,[tex]\rm V_P = 12 \ v[/tex]

Secondary coil voltage,[tex]\rm V_s = ?[/tex]

No turns in the primary coil,[tex]\rm N_p[/tex]

No turns in the secondary coil,[tex]\rm N_s[/tex]

From the given condition;

[tex]\rm N_s = 3N_p[/tex]

For a transformer,

[tex]\rm \frac{V_p}{V_s}= \frac{N_p}{N_s} \\\\ \rm \frac{12}{V_s}= \frac{N_p.}{3N_p} \\\\ V_s = 36 \ V[/tex]

Hence the voltage of the secondary will be 36 V.

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Considering the Watt's law, if a current uses 31 amperes and has a voltage of 110 volts, it dissipates 3,410 Watts.

Current is the physical magnitude that expresses the amount of electricity that flows through a conductor in unit time and is measured in amps.

The voltage is the difference between the electrical charge that leaves the source and the one that finally reaches the end of the circuit. It is expressed in volts.

Watt's Law refers to the electrical power of an electronic component or device and is defined as the power consumed by the load is directly proportional to the voltage supplied and the current flowing through it. The unit of power is the Watt.

Knowing the voltage and current, this law is expressed as:

P = V×I

In this case, you know:

Replacing in Watt's Law:

P = 110 volts× 31 amperes

Solving:

P= 3,410 Watt

Finally, if a current uses 31 amperes and has a voltage of 110 volts, it dissipates 3,410 Watts.

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Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.

The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.

The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.

The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.

Hence option B is corect.

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Answer:

Pressure will triple   (Boyle's Law)  

Explanation:

Assuming a constant temperature:  (compressing gases usually raises the temp significantly):

P1V1 = P2V2

P1V1 / V2   = P2      

   Now change V2  to 1/3 V2:

P1V1 / (1/3 V2 ) = P2 / (1/3 )

P1V1/ (1/3 v2 ) = 3 P2                <======= THE PRESSURE WILL TRIPLE

    This is Boyle's Law

The magnitude of the net force that the atmosphere applies to the roof when the outside pressure drops suddenly is  1.9 x 10⁵ N.

The pressure is the amount of force applied per unit area.

Pressure p = Force/Area

Given is a house has a roof (colored gray) with the dimensions shown in the drawing. The outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can adjust.

The pressure difference ΔP = 13.2 mm of Hg

The length of the roof l = 14.5m

the breadth of the roof h = 4.21m

The force exerted by pressure is

Force, F = P x A

         = (13.2 mm of Hg) [(133 N/m²) /1 mm of Hg ](14.5 x 4.21)

         = 107,170.6 N

Then the net vertical force

    Fnet = 2F cos30

    Fnet = 2 ( 107,170.6) cos30

    Fnet = 185625 N

    Fnet =1.9 x 10⁵ N

The direction of the force is downwards, since the horizontal components of the forces cancel each other.

Hence,  magnitude of the net force is 1.9 x 10⁵ N

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Answer:

Explanation:

There are different ways to investigate the factors that affect resistance. In this practical activity, it is important to:record the length of the wire accuratelymeasure and observe the potential difference and currentuse appropriate apparatus and methods to measure current and potential difference to work out the 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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