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iogann1982 [59]
3 years ago
15

So, like my previous question, how do you solve for 4/5 - 1/2?

Physics
2 answers:
andreyandreev [35.5K]3 years ago
7 0

Answer:

I believe its  x = 22

ZanzabumX [31]3 years ago
6 0

Answer:

3/10

Explanation:

4/5(2)

-1/2(5)

8/10-5/10

3/10

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What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af
sladkih [1.3K]

Answer:

L=0.0045\ kg-m^2/s

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, \omega=4300\ rpm=450.29\ rad/s

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

L=I\omega

Where I is moment of inertia

For sphere, I=\dfrac{2}{5}mr^2

L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s

So, the magnitude of the angular momentum of the sphere is 0.0045\ kg-m^2/s.

4 0
2 years ago
Matter can undergo chemical reactions and nuclear reactions, which
asambeis [7]

Answer:

C

Explanation:

they involve breaking and making chemical bonds

8 0
1 year ago
Why does a third class lever cannot magnify force?​
maw [93]

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

5 0
3 years ago
John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

3 0
3 years ago
HELP...<br> 3.00 amu = _____ Mev.<br><br> 3.22 x 10-3<br> 2.79 x 103<br> 3.10 x 102
DanielleElmas [232]
The correct answer is:
2.79 \cdot 10^3 MeV
Let's see why.

1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
3 amu = 3 \cdot 934 MeV  \sim 2790 MeV = 2.79 \cdot 10^3 MeV
5 0
3 years ago
Read 2 more answers
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