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attashe74 [19]
3 years ago
8

When it reacts with another element, an oxygen atom gains two electrons.

Physics
2 answers:
KatRina [158]3 years ago
7 0

Answer:A

Explanation:

luda_lava [24]3 years ago
3 0

Answer:

A

Explanation:

An ion with a negative charge

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Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction
andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

3 0
3 years ago
A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.
JulijaS [17]

Answer:

D) 61 beats per minute

Explanation:

5 0
3 years ago
Question is attatched!
lina2011 [118]
All it does is lets him pull in a more convenient direction to raise the load. It has no effect on the required force.
4 0
3 years ago
A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m
belka [17]

Answer:

Energy stored in the capacitor is U=2.7\times 10^{-5}\ J        

Explanation:

It is given that,

Charge, q=1.5\ \mu C=1.5\times 10^{-6}\ C

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

U=\dfrac{1}{2}q\times V

U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36  

U = 0.000027 J

U=2.7\times 10^{-5}\ J

So, the potential energy is stored in the capacitor is U=2.7\times 10^{-5}\ J. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
4 0
4 years ago
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