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8090 [49]
3 years ago
5

What golf skill should you be working on as a beginner golfer ?

Physics
2 answers:
Simora [160]3 years ago
8 0
Getting the ball into the cup without using your hands or feet.
Leokris [45]3 years ago
7 0
Swinging needs to be the first skill mastered
You might be interested in
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
3 years ago
If we increase the force applied to an object, and all other factors remain the same, the amount of work will
soldier1979 [14.2K]
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.

-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.

-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.

-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic. 
3 0
3 years ago
Read 2 more answers
6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employe
MariettaO [177]

Answer:

7.5 J

Explanation:

To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:

Mass (m) = 15 Kg

Velocity (v) = 1 m/s

Energy (E) =?

E = ½mv²

E = ½ × 15 × 1²

E = ½ × 15 × 1

E = ½ × 15

E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

3 0
3 years ago
Under what conditions can an ideal gas undergo a change of state without doing external work?
Cerrena [4.2K]
Nswer 

<span>Work done = Pressure * ΔV [ change in volume ] </span>

<span>If ΔV=0, then no work is done </span>

<span>Zero work thermodyanamic process is called ' isochoric process.' </span>

<span>For example if a gas heated in a rigid container: the pressure and temperature </span>
<span>of the gas will increase, but the volume will remain the same. </span>

<span>This is called an isochoric thermodynamic process. </span>

<span>It is actually possible to do work on a system without changing the </span>
<span>volume, as in the case of stirring a liquid</span>
8 0
3 years ago
Two point sources, vibrating in phase, produce an interferencepattern in a ripple tank. If the frequency is increased by 20%, th
Vera_Pavlovna [14]

Answer:

option A

Explanation:

given,

frequency is increased by 20%

we know,

\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda...........(1)

where

x_n is the perpendicular distance between the point the interference pattern is obtained,

L is the distance between the center of the two point sources

and λ is  the wavelength of light.

If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.

From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.

Hence, the correct answer is option A

8 0
3 years ago
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