Answer:
4.96×10¯¹⁰ N
Explanation:
The following data were obtained from the question:
Mass 1 (M1) = 300 Kg
Mass 2 (M2) = 300 Kg
Separating distance (r) = 110 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Gravitational force (F) =?
The gravitational force between the two goal posts can be obtained as follow:
F = GM1M2 / r²
F = 6.67×10¯¹¹ × 300 × 300 / 110²
F = 6.003×10¯⁶ / 12100
F = 4.96×10¯¹⁰ N
Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N
To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:
Mass of copper wire
Density
Resistively of copper
Resistance (R) = 0.390\Omega
Volume is defined as,
(1)
We know that,
(2)
Multiplying equation we have
Therefore the length of the wire is 1.68m
D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.
Answer:
Detailed step wise solution is attached below
Explanation:
(a) wavelength of the initial note 2.34 meters
(b) wavelength of the final note 0.389 meters
(d) pressure amplitude of the final note 0.09 Pa
(e) displacement amplitude of the initial note 4.78*10^(-7) meters
(f) displacement amplitude of the final note 3.95*10^(-8) meters
<span>Answer:
Let m = mass of cannon
Then
10000 = ma
a = 10000/m
v^2 = u^2 + 2as
v^2 = 0 + 2as
84^2 = 2(2.21)(10000/m)
84^2 m = 4.42(10000)
m = 6.264172336
= 6.26 kg
Part 2
Range = u^2sin(2x38)/g
= 84^2sin(76)/9.8
= 698.6129229
= 698.6 m</span>