Answer:
The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian.
From any location on Earth you see only half of the celestial sphere, the half above the horizon.
If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90° from the celestial poles, would lie along your horizon.
Answer:
The unit for power is the Watt
Answer: 4.9 x 10^6 joules
Explanation:
Given that:
mass of boulder (m) = 2,500 kg
Height of ledge above canyon floor (h) = 200 m
Gravita-tional potential energy of the boulder (GPE) = ?
Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).
GPE = mgh
GPE = 2500kg x 9.8m/s2 x 200m
GPE = 4900000J
Place result in standard form
GPE = 4.9 x 10^6J
Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules
Answer:
The maximum mass the bar can support without yielding = 32408.26 kg
Explanation:
Yield stress of the material (
) = 200 M Pa
Diameter of the bar = 4.5 cm = 45 mm
We know that yield stress of the bar is given by the formula
Yield Stress = 
⇒ = 
---------------- (1)
⇒ Area of the bar (A) = 
×
⇒ A = × 
⇒ A = 1589.625 
Put all the values in equation (1) we get
⇒ 
= 200 × 1589.625
⇒ = 317925 N
In this bar the is equal to the weight of the bar.
⇒ = 
× g
Where is the maximum mass the bar can support.
⇒ = 
Put all the values in the above formula we get
⇒ = 
⇒ = 32408.26 Kg
There fore the maximum mass the bar can support without yielding = 32408.26 kg
Answer:
the time needed for her to close the door is 1.36 s.
Explanation:
given information:
Force, F = 220 N
width, r = 1.40 m
weight, W = 790 N
height, h = 3.00 m
angle, θ = 90° = π/2
to find the times needed to close the door we can use the following equation
θ = ω₀t + 1/2 αt²
where
θ = angle
ω = angular velocity
α = angular acceleration
t = time
in this case, the angular velocity is zero. thus,
θ = 1/2 αt²
now, we can find the angular speed by using the torque formula
τ = I α
where
τ = torque
I = Inertia
we know that
τ = F r
and
I = 1/3 mr²
so,
τ = I α
F r = 1/3 mr² α
α = 3 F/mr
= 3 F/(w/g)r
= 3 (220)/(790/9.8) 1.4
= 5.85 rad/s²
θ = 1/2 αt²
π/2 = 1/2 5.85 t²
t = 1.36 s
