Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.
Let V = the volume of the sample.
Then
W = ρVg
or
V = W/(ρg)
= (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
= 4.4972 x 10⁻⁵ m³
Answer: The volume is 44.972 x 10⁻⁶ m³
Answer:
45° to min.
1° ----> 60 min
45° ----> ?.
45 * 60
2700 min
<h3><u>2</u><u>7</u><u>0</u><u>0</u><u> </u><u>min</u></h3>
Answer: i didnt look at your question its to long
Explanation:
Answer:
a) The distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object is 5 m/s².
Explanation:
a) The distance can be found using the equation of gravitational force:

Where:
G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²
M: is the Earth's mass = 5.97x10²⁴ kg
m: is the object's mass = 0.4 kg
F: is the force or the weight = 2.0 N
r: is the distance =?
The distance is:
Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.
b) The initial acceleration of the object can be calculated knowing the weight:
Where:
W: is the weight = 2 N
a: is the initial acceleration =?

Therefore, the initial acceleration of the object is 5 m/s².
I hope it helps you!
Answer:
3.0 x 10¹ Nm
Explanation:
Torque = F x r
Where F is force applied and r is perpendicular distance from pivot point . r
is also called lever arm
Here F = 15 N and r = 2.0 m
Torque
= 15 N X 2.0 m
= 3.0 10¹ Nm.