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3 years ago

A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.

2 answers:

34kurt3 years ago

6 0

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

$F = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

Gnom [1K]3 years ago

4 0

Answer:

all of these

Explanation:

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Which part of the battery is electron enriched?

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The answer is b. Negative terminal.

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3 years ago

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The chart below shows the average surface temperature or temperature range for each of the eight planets.

Vaselesa [24]

Answer:

A) Earth and the other inner planets have higher average surface temperatures than the outer planets.

Explanation:

the earth and the other inner planets have higher average surface temperatures than the outer planets.

The reason for this response is due to the distance between the sun and the respective planet, the source of energy generation is the sun and the only way in which the temperature increase of each planet is guaranteed is by radiation, the further away a planet is from its star, its temperature will decrease. Although it is also important to highlight the atmospheric composition of the planet if this planet in its stratosphere has high density clouds that do not allow the entry of solar radiation, the temperature of the planet's surface will not increase, independent of the distance from the sun, but these are more complex cases where specialists in that area enter to perform a study in detail.

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3 years ago

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The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the

ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0

3 years ago

An engineer can increase the magnitude of the magnification of a compound microscope by

jok3333 [9.3K]

Answer:

Option B

Explanation:

Magnification of Microscope is

$M = M_o \times M_e$

Mo= Magnification of objective lens and

Me= magnification of the eyepiece.

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.

Magnification,

$M\ \alpha\ \dfrac{1}{f}$

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B i.e. using shorter focal length

6 0

3 years ago

Now, using your mass (in kg), and the figures for g (in the table below), you can calculate your weight on other planets.

Licemer1 [7]

Answer:

1) Weight on Mercury

$F =W=mg=68.11 \times 3.61 m.s^{-2}$

Explanation:

do the same to the rest and use your calculator to find the weight in N.

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3 years ago