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2 years ago

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

Physics

2 answers:

Simora [160]2 years ago

8 0

Getting the ball into the cup without using your hands or feet.

Leokris [45]2 years ago

7 0

Swinging needs to be the first skill mastered

You might be interested in

eimsori [14]

Explanation:

$d = \frac{1}{2} a {t}^{2} + vt \\ d = 75 \\ t = 5 \\ v = 0 \\ \\ 75 = \frac{1}{2} \times 25a + 0 \\ a = 6 \frac{m}{ {s}^{2} } \\$

7 0

2 years ago

The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tis

Eduardwww [97]

Answer:

m' = 1 x 10⁻⁶ kg/s

Explanation:

Given that

Diffussion constant = 1 x 10⁻¹¹

Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

$m'=\dfrac{m}{2}$

$m'=\dfrac{2\times 10^{-6}}{2}\ kg/s$

m' = 1 x 10⁻⁶ kg/s

6 0

2 years ago

The magnitude of a negative of a vector is negative. O True O False

Paha777 [63]

Answer:

The given statement is false.

Explanation:

For any negative vector

$\overrightarrow{r}=-x\widehat{i}-y\widehat{j}$

The magnitude of the vector is given by

$|r|=\sqrt{(-x)^{2}+(-y)^{2}}\\\\|r|=\sqrt{x^2+y^2}$

As we know that square root of any quantity cannot be negative thus we conclude that the right hand term in the above expression cannot be negative hence we conclude that magnitude of any vector cannot be negative.

8 0

2 years ago

Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. You (inertia 80 kg ) then move across horizontal sn

RSB [31]

Answer:

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128 kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

5 0

2 years ago

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Naya [18.7K]

Answer:

$2.966\times 10^{-11}\ N$

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = $6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}$

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

$F=\dfrac{GMm}{d^2}$

Plug in the given values and solve for 'F'. This gives,

$F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})$

The same force is experienced by both cannonball and marble.

Therefore, the gravitational force of the marble is $2.966\times 10^{-11}\ N$

3 0

2 years ago