Determine the amount of power
used in holding a 25 kg box, 1.5
meters above the floor, for 60
seconds.

[?] W

(answer is not 6.13)

Thank you in advance!

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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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 x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

There are three little objects that are densely spaced out in the x-y plane.;

The value of the masses

m₁=1.41 kg

m₂=2.55 kg

m₃=3.19 kg

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3x_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 6 +3.19 \times 4}{1.41+2.55+3.19} \\\\ X_{cm} =4.31 \ m[/tex]

[tex]\rm Y_{cm} =\frac{ (W_1y_1 + W_2y_2 + W_3y_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 4 +3.19 \times 7}{1.41+2.55+3.19} \\\\ X_{cm} =3.12 \ m[/tex]

Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.

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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

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.

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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D.

The gravitational of a body is possessed by the body due to the virtue of its position.

The formula for gravitational potential energy is,

                                P.E = mgh joules

Substituting the values

196*9.8*250= 49k

Answer:

0.4

Explanation:

ω = km

ω = 1000 x 4

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

Answer:

r = 2.0m

when t = 0, the angular displacement is zero.

Explanation:

When the angular acceleration is 0.01 m/s2, you can calculate

angular acceleration = change in angular velocity/ time

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

Answer:

See below

Explanation:

The work done equals the increase in Potential Energy

PE = mgh

    = 150 * 9.81 * 1 = 1471.5 j

3 sets of 10 is 30 times this = 44145 j

The amount of work done by a football player in the weight room when he squats 150 kg will be equal to 1471.5 J.

Potential energy is a form of stored energy that is dependent on the relationship among different components. When a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the floor as opposed to falling to the ground, it has more potential energy. It is capable of carrying out additional work when raised.

Potential energy is a feature of systems rather than of particular bodies or particles; for instance, the system created up of Earth and the elevated ball has more energy stored as they become further apart.

Potential energy develops in systems components whose configurations, or spatial arrangement, determine the amount of the forces they apply to one another.

As per the instructions given in the question,

Work performed is equivalent to Potential Energy Increase,

PE = mgh

= 150 × 9.81 × 1

PE = 1471.5 J

Also,

3 sets of 10 are 30 times this = 44145 J

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Answer:

di

Explanation:

The prefix is di. CO2 is carbon dioxide. di means 2.

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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(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (2ati + bj) m/s and a = 2ai m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 4 m/s².

x = at²i + btj

x = 2t²i + tj

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

x = at²i + btj

v = dx/dt

v =  (2ati + bj) m/s

a = dv/dt

a = 2ai m/s²

a = 2ai m/s²

a = 2(2)(1)

a = 4 m/s²

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Answer:

Tolerate

Explanation:

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

The internal energy of the gas after the adiabatic compression will be 30.398 × 10⁶ J

When energy is moved from one store to another, work is completed. Work done on the gas is taken as -ve.

Given data;

pressure(P)=3.0 atm = 303975 N/m²

The initial volume, V₁

work is done on the gas., W=?(-ve)

Change in heat, ΔQ=0

Change in the internal energy of the gas., ΔE

The work done on the gas;

W = -PΔV

W= - 303975 N/m² × 100 cm³

W = - 30.3 × 10⁶ J

The internal energy is found as;

ΔE=q+w

ΔE= 0-30.3 × 10⁶ J

ΔE= -30.3 × 10⁶ J

E₂-E₁= -30.3 × 10⁶ J

E₂ = -30.3  × 10⁶ J +500

E₂ = 30.398 × 10⁶ J

Hence, the internal energy of the gas after the adiabatic compression will be 30.398 × 10⁶ J

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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.

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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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:

Joes displacement is 54m

Answer:

B. Outside the nucleus.

Explanation:

Electrons orbit the nucleus of the atom.

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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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

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.