For Blake:
3 boxes at a distance of 10 meters each, each box weighs 20 N
Work done by Blake = 3 * 10m * 20N
= 600 J
Power = 600 J/ 2 min
= 300 J/min
For Sandra:
4 boxes, 15 N each at a distance of 12 meters each.
Work done by Sandra = 4 * 15 N *12m
= 720 J
Power = 720 J/ 4 min
= 180 J/min
Blake does less work than Sandra.
Blake's power is more than Sandra's.
<span>The unknown substance is silver.
I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so
273 g / 26 mL = 10.5 g/mL
Looking up a list of elements sorted by density, I see the following:
10.07 Actinium
10.22 Molybdenum
10.5 Silver
11.35 Lead
And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>
Answer:
4.4 seconds
Explanation:
Given:
a = -5.5 m/s²
v₀ = 0 m/s
y₀ = 53 m
y = 0 m
Find: t
y = y₀ + v₀ t + ½ at²
0 = 53 + 0 + ½ (-5.5) t²
0 = 53 − 2.75 t²
t = 4.39
Rounded to two significant figures, it takes 4.4 seconds for the object to land.
According to your formula, because Th>Tc, the density of hot air greater than the density of cold air.<span>Hint: You can rearrange the ideal gas law to get </span>
<span>Reference https://www.physicsforums.com/threads/density-of-hot-air.528991/</span>
Answer:
<em>The object with the twice the area of the other object, will have the larger drag coefficient.</em>
<em></em>
Explanation:
The equation for drag force is given as
where 
IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
= coefficient of drag
A = area of the object
Note that is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
∝ 
A
where is the pressure exerted by the fluid on the area A
Also note that = 
which also clarifies that the drag force is approximately proportional to the abject's area.
<em>In this case, the object with the twice the area of the other object, will have the larger drag coefficient.</em>
