1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Amanda [17]
3 years ago
15

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Physics
1 answer:
Olenka [21]3 years ago
8 0

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. So we can use the following suvat equation:

v=u+at

where

v is the  final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

a=g=9.8 m/s^2 is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

v=0+(9.8)(0.75)=7.35 m/s

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

You might be interested in
In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
La masa de un camión es de
Nutka1998 [239]

Answer:

45000 kg and 45 tons  

Explanation:

The expression in kilograms and tons is shown below;

As we know that

1 gr is 0.001 kg

So, 45000000 = 45,000 kg

And,

1 kg = 0.001 tons  

So, 45000 kg = 45 tons    

Therefore the same would be considered

7 0
3 years ago
Graphs often permit scientists to make estimations for items when there is no data. Please select the best answer from the choic
Zinaida [17]

Answer:

no they don't but I think it's 8

3 0
2 years ago
How do you find the net force acting on an object?
weqwewe [10]

Answer:

C. Add all the force vectors

Explanation:

The net force acting on an object is the vector sum of all the  forces on the object.

Remember, Newton's first law tells us a body at rest will remain at rest or that in uniform motion will continue in motion unless acted by unbalanced forces.These unbalanced forces act in all direction towards the body thus to get the net force you require a summation of all these force with respect to their magnitudes and directions.

For example a force of 3N towards the East direction acting on a body and another force of 2N towards the West direction on the same body will generate a net force of 1N towards the East direction.

4 0
3 years ago
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i
Sergio039 [100]
<span>We can use an equation to find the gravitational force exerted on the HST. F = GMm / r^2 G is the gravitational constant M is the mass of the Earth m is the mass of the HST r is the distance to the center of the Earth This force F provides the centripetal force for the HST to move in a circle. The equation we use for circular motion is: F = mv^2 / r m is the mass of the HST v is the tangential speed r is the distance to the center of the Earth Now we can equate these two equations to find v. mv^2 / r = GMm / r^2 v^2 = GM / r v = sqrt{GM / r } v = sqrt{(6.67 x 10^{-11})(5.97 x 10^{24}) / 6,949,000 m} v = 7570 m/s which is equal to 7.570 km/s HST's tangential speed is 7570 m/s or 7.570 km/s</span>
6 0
3 years ago
Read 2 more answers
Other questions:
  • What are the two main classifications of matter??
    9·1 answer
  • Which best explains why the pencil seems bent?
    10·1 answer
  • Defination of thrust
    13·2 answers
  • If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini
    14·1 answer
  • 4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with
    13·1 answer
  • A rhinoceros beetle rides the rim of a disk rotating like a merry-go-round counterclockwise.
    15·1 answer
  • What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl
    15·1 answer
  • Why do things become hot or cold?<br> Relate to Thermal Energy and The Law of Conservation of Energy
    11·1 answer
  • How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers
    12·1 answer
  • Sheila uses a 45N force on her bowling ball across a 15m lane. What work did she do on the bowling ball? Show your work.​
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!