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2 years ago

PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!

Physics

2 answers:

ra1l [238]2 years ago

3 0

Answer:

Explanation:

This is true because culture in Asia focus more on energy that is why we see kung fu and taekwondo etc.

zaharov [31]2 years ago

3 0

Answer:

I say it might be true points go to the person above me if I get it right

Explanation:

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A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic

ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

$I_i$ = Initial angular momentum = $275\ kgm^2$

$\omega_i$ = Initial angular velocity = 14 rev/min

$I_f$ = Final angular momentum

$\omega_f$ = Final angular velocity

Here the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s$

The final angular velocity is 0.2932 rad/s

7 0

3 years ago

The rebirth of science and learning during the fifteenth century is termed the

Pepsi [2]

The Renaissance, was the rebirth to science as well as many other advancements

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3 years ago

Consider an alien on a planet with an acceleration of gravity equal to 20 m/sec^2. If the alien's mass is 10 kg, how much does t

mr_godi [17]

Answer:

the weight of alien is 200 newtons

Explanation:

The computation of the alien weight is shown below:

Given that

Acceleration = 20m/sec^2

And, the mass is 10 kg

So, the weight of alien is

= 20 × 10

= 200 newtons

hence, the weight of alien is 200 newtons

6 0

2 years ago

What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.

Luden [163]

Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.

8 0

3 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers