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3 years ago

Helpppppppppppp meeeeeeeeeeeeeeeee

Physics

2 answers:

Trava [24]3 years ago

6 0

kinetic energy to gravitational potential energy

Irina18 [472]3 years ago

6 0

Answer:

I don't know if I'm correct but it might be

The second option with kinetic energy.

And again idk if im correct.

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A 2-ft-diameter hemispherical plexiglass "bubble" is to be used as a special window on the side of an above-ground swimming pool

Taya2010 [7]

Answer:

a) 1156.8 lb

b) 130.7 lb

Explanation:

See attachment

6 0

3 years ago

A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is

abruzzese [7]

Answer:

As the falcon begins to fall towards the ground, then there will be two forces acting on it, 1) weight of the falcon acting downward and 2) the drag force acting on it due to it's motion through the air.

Explanation:

7 0

2 years ago

A stagehand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N. W

Step2247 [10]

Answer:

Option (a)

Explanation :

A stage hand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N.

Hence Force applied $\text { Fapplied }=176 \mathrm{N}$

Force on piece of scenery $\mathrm{F}_{\mathrm{g}}=490 \mathrm{N}$

$\Sigma \mathrm{F} \mathrm{y}=\mathrm{F} \mathrm{n}-\mathrm{Fg}=0$

$\mathrm{Fn}=\mathrm{Fg}$

$\mathrm{FS}_{\mathrm{max}}=\mathrm{F}_{\mathrm{applied}}$

µk = $\frac{\mathrm{Fs} \max }{\mathrm{Fn}}$

= $\frac{\text { Fapplied }}{\mathrm{Fg}}$

= $\frac{176}{490}$ =0.36

coefficient of static friction is 0.36

4 0

3 years ago

A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

pentagon [3]

Answer:

The block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

$v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}$

Now, plug in the given values and solve for 'a'. This gives,

$a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2$

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

$v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}$

Now, plug in the given values and solve for 'S'. This gives,

$S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m$

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0

3 years ago

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

Sever21 [200]

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

5 0

3 years ago