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
Tanya [424]
3 years ago
8

Thalia is drafting a plan to move a large, perfect sphere concrete sculpture that is in front of her office building. Describe t

he considerations Thalia would need to make if she was drafting a plan to move the sculpture
Physics
2 answers:
svet-max [94.6K]3 years ago
7 0

Sample Response: Thalia would need to consider that the sculpture's inertia is keeping the object in place. Newton's first law applies to this situation. The inertia would have to be overcome to enable the sculpture to move, but the inertia would also keep the sculpture moving after its movement was started. The force that is required to move the sculpture would depend on the mass and acceleration at which Thalia wanted the sculpture to move. Finally, any force applied to the sculpture would be resisted by the sculpture in an equal and opposite manner.

Charra [1.4K]3 years ago
3 0

Answer:She would need to first know the weight of the sculpture and what she is going to move it with then she will need to use newton's second law to calculate the amount of force needed to move it

Explanation: I just did the assignment on edgunity

You might be interested in
Write a paragraph of no less than five sentences explaining how Newton's First Law of Motion supports people's need to wear seat
Inessa [10]
Newton's first law can be taken to mean that if something is moving it tends to keep moving. if at rest it tends to stay at rest.

so, in a car, you and the car are both moving, say at constant speed. Now you're not actually connected to the car as in clamped to it, not yet at least. You're simply sitting in it at rest with respect to it.

but, someone slams on the brakes for whatever reason. The car slows down/stops. what do you do ? well, you would keep going. and moving a few feet in a car can be dangerous, esp if you're moving at high speed. Unless of course you're clamped to the seat, and the seat is clamped to the car and the car is clamped together. then when the car brakes, yes you'll feel the braking effect, but the belt will restrict your movement, keeping you safe, if shocked and bruised.
7 0
3 years ago
The most abundant state of matter in the universe is also the state with the most kinetic energy. That is
ella [17]

C)

Plasma is the most abundant state of matter in the universe and has the most kinetic energy. It can be found in all the planets of our solar system, the sun, and other stars

8 0
3 years ago
A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
3 years ago
What is 3,278,000 as scientific notation?
jonny [76]
3.278*10^6 I think. Sorry if it’s wrong.
4 0
3 years ago
Can someone help me please??
Mrrafil [7]

1. Our solar system is the only place in the universe where gravity played a key part in the formation of planets.

2. Rocky planets are small, dense, and orbit relatively close to the sun, compared to the Jovian planets, which are large, less dense, and orbiting far from the sun.

3. _______

7 0
3 years ago
Other questions:
  • Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this stil
    14·1 answer
  • why does light look colorless and how is made (I know this just Wana see if anyone will get it correct )
    12·1 answer
  • If you have 500 g of water at 25oc and wish to heat it to 74o c, what is the specific heat of water
    14·1 answer
  • Power is __________________. Power is __________________. the work done by a system the force required to push something the spe
    11·1 answer
  • Which kind of electromagnetic waves has a wavelengths greater than 10 cm?
    14·1 answer
  • Two runners start simultaneously from the same point on a circular 200-m track and run in the same direction. One runs at a cons
    6·1 answer
  • Two objects each with a mass of 5x10^15 kg have a gravitational
    6·1 answer
  • What is one characteristic of the organization of the Periodic Table?
    12·1 answer
  • Do you think it is possible to control the magnetic properties of a magnet? Can a magnet be turned on and off? Explain your answ
    13·1 answer
  • If you drop an object, it will accelerate downward at a rate of 9.8 meters per second per second. if you instead throw it downwa
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!