Answer:
1.31×10¯⁶ N
Explanation:
From the question given above, the following data were obtained:
Mass of John (M₁) = 81 Kg
Mass of Mike (M₂) = 93 Kg
Distance apart (r) = 0.620 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The gravitational force between the two students, John and Mike, can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 81 × 93 / 0.62²
F = 6.67×10¯¹¹ × 7533 / 0.3844
F = 1.31×10¯⁶ N
Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N
Answer:
8.36e2
Explanation:
Use a scientific calculator
I think the smartest answer would be C sorry if i’m writing though
Answer:
Gold
Explanation:
Given:
Mass of sample = 63.5 g
Mass of water = 60.2 g
Find:
Object
Computation:
Mass of water displaced = 63.5 g - 60.2 g
Mass of water displaced = 3.3 g
So, volume in water = 3.3 cm³
Density = Mass / Volume
Density = 63.5 g / 3.3
Density = 19.24
So,
Object ,must be gold.
The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant = 8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg
Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
