Given: Normal pull of gravity g = 9.8 m/s²;
g = 0.855 m/s² (at a certain distance)
Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²
Mass of the Earth Me = 5.98 x 10²⁴ Kg
Radius r = ?
g = GMe/r²
r = √GMe/g
r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)
r = 2.16 x 10⁷ m or
r = 21,610 Km
.
<h2>Answer:</h2>
Acceleration= -1.11 m/sec²
<h2>Explanation:</h2>
Date Given to us is
Mass = 150 kg
Time = 1.5 minutes = 90 seconds
Distance = 2500 meters
Initial Velocity = 120 m/s
Final Velocity = 20 m/s
Acceleration = ?
<u>Solution:</u>
By using First Equation of motion
Vf = Vi + at
Putting the values
20 = 120 + a (90)
Subtracting 120 on both sides
20-120 = 120 + a(90) - 120
-100 = 90 a
Dividing both sides by 90
-100/90 = 90 a / 90
-1.11 = a
So the acceleration is -1.11 m/s²
Answer:
τ = 32.8635 N-m (counterclockwise)
Explanation:
Given
M = 2 kg
L = 2 m
r = 0.10 m
m₁ = 4 kg
r₁ = (1.00-0.30)m = 0.70 m
m₂ = 5 Kg
r₂ = (0.90-0.75)m = 0.15 m
In order to determine the torque on the meterstick if it is received and allowed to pivot about the 90 cm mark (ypu can see the pic to understand the question), we apply:
τ = r₁*m₁*g + r₂*m₂*g - r*M*g = g*(r₁*m₁+r₂*m₂-r*M)
⇒ τ = (9.81 m/s²)(0.70 m*4 kg + 0.15 m*5 Kg - 0.10 m*2 kg)
⇒ τ = 32.8635 N-m (counterclockwise)
Answer:
m = 63 grams
Explanation:
ω = 10 cycles/s(2π radians/cycle) = 20π rad/s
ω = √(k/m)
m = k/ω² = 250/(20π)² = 0.06332... kg
