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irga5000 [103]
3 years ago
6

What is the volume of an object that has Mass=100g and Density= 745g/mL.

Physics
1 answer:
lana66690 [7]3 years ago
6 0

Answer:0.13ml

Explanation:

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A mid-ocean ridge under the ocean is MOST similar to a _____ on land.
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A mid ocean ridge is a under water mountain range, knowing this what would you say it is most similar to on land? hope this helped!
5 0
3 years ago
HELP
dem82 [27]

Answer:

Explanation:

28 / 70 = 0.3857142...  = 0.39 hr

280 / 100 = 2.8 hrs.

(100 - 0) / 10 = 10 m/s²

(60 - 20) / 4 = 10 m/s²

3 0
3 years ago
Number 3 How to do?​
Nat2105 [25]

Answer:

1 m/s

Explanation:

Impulse = Change in momentum

Force × Time = Mass(Final velocity) - Mass(Initial Velocity)

(1.0)(1.0) = (1.0)(Final Velocity) - (1.0)(0)

Final velocity = <u>1 m/s</u>

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4 years ago
What is the difference between: first moment of area and second moment of area?
zhannawk [14.2K]
To determine the centroid of the object first moment of area is used.

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6 0
3 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
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