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3 years ago

2. What is the standard value of acceleration due to gravity or "g"?

Physics

2 answers:

lawyer [7]3 years ago

8 0

Answer:

9.80665 m/s^2

Explanation:

This is the value on Earth.

svp [43]3 years ago

6 0

Answer:

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2 (about 32.17405 ft/s2).

Explanation:

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Debbie places two shopping carts in a cart corral. She pushes the first cart, which then pushes a second cart. What force is bei

ExtremeBDS [4]

Hey there,

Your question states: Debbie places two shopping carts in a cart corral. She pushes the first cart, which then pushes a second cart. What force is being exerted?

based by looking at this statement above about Debbie, I understand that she (pushed) Cart (A) first. And then, she exerted (Cart B) next. From the option's that are listed above, I only see two. But from my own words, not from the only (two) options above, I see that (Debbie first exerts the second cart on to the first cart). This reason would be because the first cart is already in the corral. So then she would put the second one in there, this would mean that the second one would push the other one in there. Which means that the velocity would also be in half.

I hope you grabbed my answer in there.
~Jurgen

5 0

3 years ago

Read 2 more answers

The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th

Yuri [45]

Answer:

$R=64.32\ lb\\\\\theta=84.3\°$

Explanation:

Given:

Ratio of lift force to drag force is, $\frac{L}{D}=10$

Lift force on a short section is, $L=64\ lb$

Magnitude of resultant, $R= ?$

The angle of 'R' with the horizontal is, $\theta=?$

We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.

For calculating 'R', we first compute drag force 'D'.

As per question:

$\frac{L}{D}=10\\\\D=\frac{L}{10}=\frac{64\ lb}{10}=6.4\ lb$

Now, the magnitude of resultant 'R' is given as:

$R=\sqrt{L^2+D^2}$

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

$R=\sqrt{64^2+6.4^2}\\\\R=\sqrt{4096+40.96}\\\\R=\sqrt{4136.96}=64.32\ lb$

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.

Now, the angle $\theta$ is given as the arctan of the ratio of the lift and drag force.

Therefore,

$\theta=\tan^{-1}(L/D)\\\\\theta=\tan^{-1}(10)\\\\\theta=84.3\°$

Therefore, the angle made with the horizontal is 84.3°.

6 0

3 years ago

Remaining Time: 1 hour, 49 minutes, 34 seconds.

ohaa [14]

Explanation:

i think C . it is twice the size of the object

4 0

3 years ago

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

Westkost [7]

Answer:

The range is 15.15 m and the time in the air is 1.01 s

Explanation:

Horizontal Motion

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$