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alina1380 [7]
3 years ago
5

Adam is teeing off on hole number two. The hole is 390 yards away. It is a par four hole. What club should he use to tee off? Ex

plain your choice.
Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

A driver.

Explanation:

Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.

You might be interested in
What is a partial and total lunar eclipse?
Minchanka [31]

Partial Lunar Eclipse:

A partial lunar eclipse is when the earth gets between the Sun and Moon. However, all three bodies are not in alignment meaning we are able to see some more like part of the moon's surface as it moves in route of the Earth's shadow.

Total Lunar Eclipse:

The three celestial bodies are perfectly aligned which allows for the earth to completely block the sun's rays from hitting/reaching the moon. The sun is positions is in back of the Earth which then causes the shadow of the earth to be cast on the Moon covering the moon completely. When that happens you get the phenomenon called a total lunar eclipse.

Hopefully this helped and good luck.



7 0
3 years ago
750 kg car zooms away from a red light with an acceleration of 7.8 m/s squared . What is the average net force in Newtons that t
natta225 [31]

Answer:

<h2>5850 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 7.8

We have the final answer as

<h3>5850 N</h3>

Hope this helps you

5 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
Can someone please explain number 8?
guajiro [1.7K]

Answer: Ax=(Vx-Vox)/(T)

Vx=Vox+Ax*T

Solving for Ax in terms of Vx, Vox, T

Vx-Vox=Ax*t

Ax=(Vx-Vox)/(T)

This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.

Any questions please feel free to ask. Thanks

8 0
3 years ago
The acceleration vector of a particle in projectile motion ________.
Alex73 [517]

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

4 0
3 years ago
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