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3 years ago

8

A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assume no air resistance. How long u

ntil the ball reaches the ground?

1 answer:

slega [8]3 years ago

8 0

The horizontal motion has no effect on the vertical drop.

From a drop, the distance the ball falls in 'T' seconds is

D = 4.9 T^2

so

2.2 = 4.9 T^2

T^2 = 2.2/4.9

T^2 = 0.449 sec^2

T = 0.67 second

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14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change

klio [65]

Lets se

And

$\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}$

$\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}$

So

$\\ \rm\Rrightarrow k\propto m$

If spring constant is doubled mass must be doubled

8 0

2 years ago

Compare the mass on Neptune (68.11) to the weight on Neptune (905.863). What is the difference? Why is there a difference?

Alchen [17]

Bbbjv. Jfuzuoud iuzTt45789

7 0

2 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago

A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and

natima [27]

The upward force exerted on the board by the support is 530.8 N.

<h3>Upward force exerted on the board by the support</h3>

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

Learn more about upward force here: brainly.com/question/6080367

#SPJ1

8 0

2 years ago

Which of the following options is correct and why?

Dimas [21]

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

6 0

3 years ago