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Leokris [45]
3 years ago
5

Why is energy required to get an object moving?

Physics
2 answers:
xxMikexx [17]3 years ago
6 0

Answer:

this is what popped up when I searched it up:In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

Explanation:

Vlada [557]3 years ago
3 0

Answer:

idk but here u go

go ez im  not in high school

Explanation:

When you start pushing or pulling a stationary object with a constant force, it starts to move if the force you exert is greater than the net forces resisting the movement, such as friction and gravity. If the object starts to move at some speed, it will acquire kinetic energy. Kinetic energy is the energy an object has because of its motion.

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An advertising balloon shaped like a giant soda can is 15 feet tall and 7 feet wide. How many cubic feet of helium will be neede
Iteru [2.4K]

Answer:

576.975 cubic feet of helium will be needed to fully inflate the balloon

Explanation:

To determine how many cubic feet of helium will be needed to fully inflate the balloon, we will determine the volume of the balloon.

From the question,

The balloon is shaped like a giant soda can.

A giant soda can is cylindrical.

Hence, we can determine the volume of the balloon shaped like a giant soda can by using the formula for finding the volume of a cylinder.

The formula for finding the volume of a cylinder is

V = πr²h

Where V is the volume of the cylinder

π is a constant (Take π = 3.14)

r is the radius of the cylinder

and h is the height of the cylinder

From the question, the balloon is 15 feet tall and 7 feet wide

Hence,

Height, h = 15 feet

Width = 7 feet

(NOTE: The width of a cylinder is the same as the diameter)

Then, diameter = 7 feet

Radius, r is given by

Radius = Diameter / 2

Then, Radius = 7 feet / 2 = 3.5 feet

∴ Radius, r = 3.5 feet

Now, for the volume of the balloon,

V = πr²h

V = 3.14 × (3.5)² × 15

V = 3.14 × 12.25 × 15

V = 576.975 cubic feet

This is the volume of the balloon.

Hence, 576.975 cubic feet of helium will be needed to fully inflate the balloon.

5 0
3 years ago
A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an
kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

  • <em>mass of the car, m = 1000 kg</em>
  • <em>speed of the car, v = 50 km/h = 13.89 m/s</em>
  • <em>banking angle, θ = 20⁰</em>

The normal force on the car due to banking curve is calculated as follows;

Fcos(\theta) = mg

The horizontal force on the car due to the banking curve is calculated as follows;

Fsin(\theta) = \frac{mv^2}{r}

<em>Divide </em><em>the second equation by the first;</em>

\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

3 0
2 years ago
During an investigation, a scientist heated 123.6 g of copper carbonate till it decomposed to form a black residue. The total ma
zloy xaker [14]

Answer:

See below explanation

Explanation:

The correspondent chemical reaction for copper carbonate decomposed by heat is:

CuCO₃ (s) → CuO (s) + CO₂ (g)

Considering all molar mass (MM) for each element ( we consider rounded numbers) :

MM CuCO₃ = 123 g/mol

MM CuO = 79 g/mol

MM CO₂ = 44 g/mol

Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.

So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which  "escaped" from our system, since it is a gas compound (CO₂)

5 0
3 years ago
What statement is true about earth tectonic plates
DaniilM [7]
Continent jig-saw shapes when puzzled and combined together, formed one big continent - Pangea, and was separated by drifts.

Fossil comparisons of different species were discovered into two different, separated continents in which when you combine them, they were one in the past.

Seismic, volcanic, and geothermal activity are found along imagined plate boundaries. 

Plates were actually rubbing against each other as evidence is seen on the formed mountain ranges.
<span>
Paleomagnetism, magnetic field placement in the layers of the rock are present.</span>
3 0
3 years ago
Read 2 more answers
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
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