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Vesna [10]
3 years ago
12

True or False. If you kick a soccer ball and there are no forces like friction acting on it will it keep moving?

Physics
2 answers:
natka813 [3]3 years ago
8 0

That's true.  It's very close to being exactly what Newton's first law of motion says.

VLD [36.1K]3 years ago
6 0

Answer:

Yes

Explanation:

Newton's laws states that gravity or force will stop the balls movement. Think of space for a example, if you pushed a soccer ball into space if would float for eternity.

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If object is above it lowest point what ever does it have
eimsori [14]
Can you reword the sentence?
6 0
3 years ago
What are the two forces that act on the orbit of a planet?
Pani-rosa [81]
A. Velocity of planet and gravity between sun and planet
7 0
3 years ago
While standing at the top of an 100 m high observatory, you accidentally dropped your phone through the grates (it is now fallin
Arturiano [62]

Using the idea of motion under gravity, it will take 0.5 seconds more before the phone hits the ground.

<h3>What is the distance covered?</h3>

The distance covered is obtained form the equations of kinematics under gravity.

Now;

v = u + gt

Recall that it was dropped from a height hence u = 0 m/s

v = gt

v = 9.8 m/s^2 * 4 s

v = 39.2 m/s

Now;

h = ut + 1/2gt^2

but u = 0 m/s

h = 1/2gt^2

h = 1/2 * 9.8 * (4)^2

h = 78.4 m

The phone  will not hit the ground within this time

For the phone to hit the ground;

h = 1/2gt^2

if h = 100 m

100 = 1/2 * 9.8 * t^2

2 * 100/9.8 =  t^2

t = √2 * 100/9.8

t = 4.5 seconds

It will take about 0.5 seconds more before the phone will hit the ground.

Learn more about motion under gravity:brainly.com/question/15120445

#SPJ1

3 0
2 years ago
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
3 years ago
In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst
dem82 [27]

Answer:

r_\alpha=16cm

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

r=\frac{mv}{qB}(1)

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\

Solving this for mv:

mv=\sqrt{2mK}(2)

Replacing (2) in (1):

r=\frac{\sqrt{2mK}}{qB}

For protons, we have:

r_p=\frac{\sqrt{2m_pK}}{eB}(3)

For alpha particles, we have:

r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)

Dividing (4) in (3):

\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm

4 0
4 years ago
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