Answer:
A 3 feet radius snowball will melt in 54 hours.
Explanation:
As we can assume that the rate of snowball takes to melt is proportional to the surface area, then the rate for a 3 feet radius will be:
T= A(3 ft)/A(1 ft) * 6 hr
A is the area of the snowballs. For a spherical geometry is computing as:
A=4.pi.R^2
Then dividing the areas:
A(3 feet)/A(1 foot) = (4 pi (3 ft)^2)/(4 pi (1 ft)^2) = (36pi ft^2)/(4pi ft^2)= 9
Finally, the rate for the 3 feet radius snowball is:
T= 9 * 6 hr = 54 hr
exothermic
Explanation:
negative means it's losing energy
The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.
To find the answer, we need to know about the orbital velocity a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbit
<h3>What's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?</h3>
- Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ m
- Orbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)
= 60.8m/s
Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.
Learn more about the orbital velocity here:
brainly.com/question/22247460
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Answer:
14cm
Explanation:
Mass per gram of the piece of wire;
2g of the wire is found in 1m
Since
100cm = 1m;
So;
100cm of the wire contains 2g of the wire
To provide 0.28g
Since;
2g of wire is made up of 100cm
0.28g of wire will be contained in 
= 14cm
14cm of the wire will contain 0.28g
Answer:
636.4 J
Explanation:
The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)
Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r
U' = 4kq²/r
= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m
= 900 Nm²/√2 m
= 636.4 J
