Part a)
in horizontal direction there is no gravity or no other acceleration
so in horizontal direction the speed of clam will remain same
Part b)
In vertical direction we can use kinematics
part c)
if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force by the worker must be equal to the friction force
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
b. The work is done on the crate by this force is the product of its force and the distance traveled s = 4.35
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
Answer:
G = 6,786 10⁻¹¹ m³ / s² kg
Explanation:
The law of universal gravitation is
F = G m M/ r²
Where G is the gravitational constant, m and M are the masses of the bodies and r is the distance from their centers
Let's use Newton's second law
F = m a
The acceleration is centripetal
a =
We replace
G m M / r² = m
G = r² / M
Let's replace and calculate
G = 2.7 10⁻³ (3.88 10⁸)² / 5.99 10²⁴
G = 6,786 10⁻¹¹ m³ / s² kg
Let's perform a dimensional analysis
[N m²/kg²] = [kg m/s² m² / kg²] = [m³ / s² kg]