If the mass of the falling object is 60kg. Calculate force. [Take g=10m/s2] ​

The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

u is the initial velocity of fall = ?  m/sec

h is the distance of fall = 200 m

g is the acceleration of free fall = 9.81 m/sec²

H is the height of the building

t is the time period = 10 second

According to Newton's second equation of motion,

[tex]\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec[/tex]

- ve shows the direction is downward.The magnitude of the initial velocity is found as;

u = 29.05 m/sec

The height of the building

[tex]\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m[/tex]

Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

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The field exerts 6.5 x 10⁸ Newton of force on EACH COULOMB of charge in the field.

We're putting 1.6 x 10⁻¹⁹ Coulomb of charge into the field.

The force on it will be

      (1.6 x 10⁻¹⁹ Coulomb) x  (6.5 x 10⁻⁸ Newton/Coulomb).

That's 1.04 x 10⁻¹⁰ Newton.

the answer is in the picture ok

good luck don't worry it's a correct

#carry on learning

You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as

What is the average force on the jet is due to the catapult?

Generally, the equation for acceleration is mathematically given as

[tex]a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2[/tex]

In conclusion, The force

F=m*a

F=15100*26.673

F=40272.3N

F 402 KN

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Answer:

i dond no can you put another question

Answer:

Given - Force = 21.4 N

Work done = 1974 J

To find - Distance

Solution -

Work done = Force * displacement

1974 j = 21.4 N * displacement

1974/21.4 = displacement

92.24

[tex]\\ \rm\Rrightarrow Work=Force\times Displacement [/tex]

[tex]\\ \rm\Rrightarrow 1974=21.4×Displacement[/tex]

[tex]\\ \rm\Rrightarrow Displacement=92.24m[/tex]

Answer:

Using 3 periods to get an accurate reading:

3T = (6.40 S - 0.90 s) = 5.50 S So, T = 1.83 s

m = 0.250 kg

Using algebra:

k= [tex]\frac{4\pi ^{2}m }{T^{2} }[/tex]=[tex]\frac{4\pi ^{2}0.250 }{1.90^{2} }[/tex]=2.95kg/sec

Explanation:

The light is bent away from the normal

Some words that have changed meaning due to technological advances are dial, type, tweet, drone, and spam.

Answer:

Tolerate

Explanation:

The weight of the object on pluto will be 15.56 N

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data;

[tex]\rm g_e[/tex] is the acceleration due to gravity on earth.

m is the mass of an object

[tex]\rm g_p[/tex] is the acceleration due to gravity on pluto.

[tex]\rm W_e[/tex] is the weight of the object on earth

[tex]\rm W_ p[/tex] is the weight of an object on pluto

The mass of the object on the earth's surface is found as;

We = m × ge

m = W / ge

m = 250 / 9.81

m = 25.51  kg

The weight of the object on pluto.

Wp = m × gp

Wp = 25.51  kg × 0.61  m/s²

Wp =15.56 N.

Hence the weight of the object on pluto will be 15.56 N

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Answer: B.

15.6 newtons

Explanation: edmentum

The height of stairs a 60 kg student would have to run in order to burn off 1 Calorie of food energy will be 7.1 meters.

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

Given data;

Mass of student,m = 60 kg

Energy,E = 4190 J

Height of stairs,h=?

The potential energy is mainly dependent upon the height of the object. The potential energy is found as;

Potential energy = mgh

E=mgh

h=E/mg

h= 4190 J / (60 kg× 9.81  m/s²)

h = 7.1 metre

Hence the height of the stairs will be 7.1 meters.

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The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

Electric field at a given distance is calculated as follows;

E = kq/r²

E₂ = (9 x 10⁹ x q)/(2²)

E₂ = 2.25 x 10⁹q

E₂ + E₀ = 0

2.25 x 10⁹q + 4 = 0

2.25 x 10⁹q = - 4

q = -4 / (2.25 x 10⁹)

q = -1.78 x 10⁻⁹

E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)

E₄ = - 1 N/C

|E₄| = 1 N/C

Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

The complete question is below:

Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?

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Answer:

it will float

Explanation:

according to Archimedes' principle, if the buoyant force is equal to the weight of the mass of the object, the object will float

Answer:

Explanation:

The amplitude is extraneous information for this question.

Period (T) is simply a measure of time (how long it takes to complete a full cycle).

The frequency (f), is the reciprocal (how many cycles it can complete per unit of time).

The frequency reported is 12Hz.  Hz is a unit meaning "cycles per second"

Thus 12cycles = 1second.

To find the number of seconds per cycle, simply create the other unit fraction and simplify:

[tex]T=\frac{1second}{12cycles}\\T=0.8\bar{3}\frac{seconds}{cycle}\\[/tex]

[tex]T=0.83seconds[/tex]

The mass in one period travels 'a distance less than 25 cm which will be 0.

The amplitude of a pendulum is the maximum displacement of the pendulum.

If the maximum displacement of the pendulum is 25 cm, the mass in one period travels a distance less that 25 cm.

Thus, then the mass in one period travels 'a distance less than 25 cm which will be 0.

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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.

Explanation:

Answer:

Approximately [tex]325[/tex] (rounded down,) assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, [tex]{\rm J}[/tex]):

[tex]\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}[/tex].

Height of the weight (should be in meters, [tex]{\rm m}[/tex]):

[tex]\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}[/tex].

Energy required to lift the weight by [tex]\Delta h = 0.2\; {\rm m}[/tex] without acceleration:

[tex]\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}[/tex].

At an efficiency of [tex]0.25[/tex], the actual amount of energy required to raise this weight to that height would be:

[tex]\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}[/tex].

Divide [tex]2.551 \times 10^{5}\; {\rm J}[/tex] by [tex]784\; {\rm J}[/tex] to find the number of times this weight could be lifted up within that energy budget:

[tex]\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}[/tex].

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

The appliance is run with a 12 V car battery and the value of the electric current will be 0.3A.

The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current.

From ohm's law;

Case 1;

V=IR

120 V = 3 A × R

R = 40 ohm

For the same appliances, the value of the resistance is the same;

Case 2;

V=IR

12 V =  I × 40 ohm

I = 0.3 A

Hence the value of the current for case 2 will be 0.3 A.

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Answer:

Explanation:

Wow, I said 'beets'..... I must be weird.

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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Hello!

We can use the following equation for magnetic force on a charged particle:
[tex]F_B = qv \times B[/tex]

[tex]F_B[/tex] = Magnetic force (N)

q = Charge of particle (1.6 × 10⁻¹⁹ C)
v = velocity of particle (5.2 × 10⁷ m/s)

B = Magnetic field strength (1.4 T)

This is a cross-product, so the equation can be rewritten to F = qvBsinφ where φ is the angle between the magnetic field and particle velocity vectors.  

Since the proton's velocity vector and the magnetic field vector are perpendicular, sin(90) = 1. We can reduce the equation to:

[tex]F_B = qvB[/tex]

Plug in the known values.

[tex]F_B = (1.6*10^{-19})(5.2*10^7)(1.4) = \boxed{1.1648 *10^{-11} N}[/tex]

Answer:

Explanation:

Comment

I think the best answer is probably A.  It's not the best reason for Earth loosing it's helium, but it is the only one that is close. The Earth derives some of its Helium (most) by the decomposition process of radio active high weight chemical that break down. Helium combines with practically nothing so once it gets moving, very little will stop it. That's the way we loose our Helium.

Still answer A.

HIV can be contracted from contact with bloodborne pathogens.

Other bloodborne diseases are HBV, malaria, syphilis and brucellosis

Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.

Blood borne pathogens can also be contacted through the following means

In conclusion; HIV can be contracted from contact with bloodborne pathogens.

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It is important to remember "association", instead of "causation" when discussing correlations because associations can be used to make predictions.

The associations among variables have a fundamental value in science because they can determine how such variables behave in hypothetical situations.

Moreover, correlation only indicates a type of association between variables and therefore it is less informative.

In conclusion, it is important to remember "association", instead of "causation" when discussing correlations because associations can be used to make predictions.

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The centripetal force is obtained by its formulas is,[tex]\rm F_C= \frac{mv^2}{r}[/tex]

The acceleration needed to move a body in a curved way is understood as centripetal acceleration.

The direction of centripetal acceleration is always in the path of the center of the course. The total acceleration is the result of tangential and centripetal acceleration.

The formula for the centripetal force is ;

[tex]\rm F_C= \frac{mv^2}{r}[/tex]

Where,

r is the radius

v is the linear speed

m is the mass of an object

Hence, the centripetal force obtained by its formulas as,[tex]\rm F_C= \frac{mv^2}{r}[/tex]

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Teacher Cellyn Kyre ensures that her classroom is always ready for the face to face classes​ is the sentence used in  quotas and interviews.

Interview is the examination of the candidate from which a person identifies the candidate is fit for the organization o r not. Maximum interview happened between the interviewer and the candidate. or sometime more people can also be present.

Thus, option D is correct.

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The average velocity is 4i m/sec

The average acceleration is 4i

Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs. T

According to the question,

x=at²i + btj ( t is in second and x is in meter).

Average velocity =  [tex]\frac{dx}{dt}[/tex]

[tex]\frac{dx}{dt} = 2at i + b j\\\\\frac{dx}{dt} = 2 * 2 * t i +1 j\\\\\frac{dx}{dt} = 4t i +j[/tex]

[tex]\frac{dx}{dt}[/tex] in the time interval of 2 to 3 sec is

at 2 sec,

[tex]\frac{dx}{dt} = 4 * 2 i +j = 8i +j[/tex]

at 3 sec ,

[tex]\frac{dx}{dt}[/tex] = 4 * 3 i + j = 12i+j

Therefore the average velocity is

[tex]\frac{(12i +j - 8i -j)}{(3-2)} = 4i[/tex]

The velocity at any given time is   [tex]\frac{dx}{dt} = 4t i +j[/tex]

The acceleration at any given time is [tex]\frac{dx^2}{dt^2}[/tex]

[tex]\frac{dx^2}{dt^2} = 4 i[/tex]

The average acceleration in the time interval  t₁=2sec and t₂=3sec

Average Acceleration =  4i

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