Response:
83%
Clarification:
At the surface, the weight can be expressed as:
W = GMm / R²
where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.
When in orbit, the weight is given by:
w = GMm / (R+h)²
where h indicates the shuttle's altitude above Earth's surface.
The weight ratio is as follows:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
For R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
Thus, the shuttle maintains 83% of its weight as it orbits.
Answer:
The convergence of light rays redirects them toward the focal point, resulting in a magnifying effect.
Explanation:
Answer:
W = 294 J
Explanation:
provided,
mass of the projectile = 2 Kg
horizontal displacement = 20 m
vertical displacement = 15 m
work performed by the gravitational force =?
the work done by gravitational force only accounts for vertical motion.
force due to gravity = m g
= 2 x 9.8 = 19.6 N
work is equal to force x displacement
W = F x s
W = 19.6 x 15
W = 294 J
Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09
Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489
Thus, the acceleration is approximately 5.5 ft/s²!!
To address this issue, we apply the de Broglie equation written as:
λ = h/mv
where h equals 6.626×10⁻³⁴ J·s
Solving for m, we substitute for v, which is 46.9 m/s:
9.74 × 10⁻³⁵ m = 6.626×10⁻³⁴ J·s / (m)(46.9 m/s)
Thus, we find that m = 0.145kg
