(D) The gravitational force between the astronaut and the asteroid.
Reason :
All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.
Answer:
0.69444 m, 0.08152 m, 0.32407 m, 0.03804 m
Explanation:
v = Velocity of sound
f = Frequency
Length of vocal tract is given by

At f = 270 Hz v = 750 m/s

At f = 2300 Hz v = 750 m/s

At f = 270 Hz v = 350 m/s

At f = 2300 Hz v = 350 m/s

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat
So here weight of the boat = Buoyancy force
let say boat is sunk by distance "h"
now we can say


now by above force balance equation we can write




so boat will sunk by total 5 mm distance
Answer:
(a) 29 cm
(b) 43.5 cm
Explanation:
(a) when loop A is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 70 cm mark
-2 N at x
Taking the sum of the torques about B:
∑τ = Iα
(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0
18 Ncm − 2 N (x − 70 cm) = 0
2 N (x − 70 cm) = 18 Ncm
x − 70 cm = 9 cm
x = 79 cm
The distance from the center is |50 cm − 79 cm| = 29 cm.
(b) when loop B is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 20 cm mark
-2 N at x
Taking the sum of the torques about A:
∑τ = Iα
(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0
-27 Ncm − 2 N (x − 20 cm) = 0
2 N (x − 20 cm) = -27 Ncm
x − 20 cm = -13.5 cm
x = 6.5 cm
The distance from the center is |50 cm − 6.5 cm| = 43.5 cm
Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.