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2 years ago

Must physics always be explained by math?

Physics

1 answer:

cluponka [151]2 years ago

5 0

Answer:

Because Physics is mainly composited of math

Explanation:

If you really think about it the only way any science works is off maths, whether that be measuring the amount of NaOH to put into a solution or determening the amount of sugar you want in a cake it all relys on maths

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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe

goldenfox [79]

Answer:

No
5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

211.1 = 80.5v²/11.5

30.16 = v² . . . . multiply by 11.5/80.5

5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

Comment on crossing the moat

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0

2 years ago

What total mass must be converted into energy

Eduardwww [97]

This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
E = m c²

 1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

 Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

 = (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

 = 1.144 x 10⁻³⁰ kg . (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:

The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

 = (2,242 x 10⁶ x 86,400 x 365) joules

 = 7.0704 x 10¹⁶ joules .

How much converted mass is that ?

 E = m c²

Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s

 Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

 = 0.786 kilogram ! ! !

THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

7 0

2 years ago

Read 2 more answers

In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th

Arlecino [84]

We first determine the vertex by using the formula,-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got

v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon

8 0

2 years ago

Why did the author most likely write "Meeting Juli a

svet-max [94.6K]

Answer:

To share a positive experience she had with a pen pal

Explanation:

Just read the story :)

8 0

2 years ago

Read 2 more answers

A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhauste

marusya05 [52]

516.154 megawatts of heat are exhausted to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ( $Q_{out}$ ), in megawatts: