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Oxana [17]
4 years ago
10

3.88 milligrams is equal to how many centigrams?

Physics
2 answers:
Ainat [17]4 years ago
7 0

Answer:

<h2>0.388 centigrams</h2>

Explanation:

Lana71 [14]4 years ago
3 0
1 centigram is equal to 10 milligrams.

Divide 3.88 milligrams by 10 to convert into centigrams.

3.88 / 10 = .388

The answer is .388 centigrams.
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Two balls have the same mass, but one has a larger volume than the other. which ball has the larger density
Aliun [14]
Answer: the ball with the less volume.
To count density you need to divide the mass by the volume. That means larger mass will be more density, but larger volume means lower density. In this case, the ball has the same mass. So the one with less volume has more density.
5 0
4 years ago
Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie
Bond [772]

Answer:

Explanation:

For parallel inductors ,

\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}

\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}

L_R=16.67 mH.

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ωL_{total} where ω is angular frequency

= 2πfL_{total}

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0
3 years ago
A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 544 m? (Round your an
Artemon [7]

Answer:

206.62313 m/s

Explanation:

u = Muzzle speed

g = Acceleration due to gravity = 9.8 m/s²

\theta = Angle at which the bullet is fired = 30°

h = Maximum height = 544 m

Maximum height is given by

h=\dfrac{u^2sin^2\theta}{2g}\\\Rightarrow u=\sqrt{\dfrac{2gh}{sin^2\theta}}\\\Rightarrow u=\sqrt{\dfrac{2\times 9.81\times 544}{sin^2(30)}}\\\Rightarrow u=206.62313\ m/s

The muzzle speed is 206.62313 m/s

6 0
4 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
Sound waves rely on matter to transmit their energy. They cannot ravel in a vacuum. True or false
UNO [17]

Answer:

false

Explanation:

8 0
3 years ago
Read 2 more answers
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