Answer:
Option E is correct.
There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.
Explanation:
Normally, ignoring air resistance, for projectile motion, the range (horizontal distance teavelled) of the motion is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile = 20 m/s
θ = angle above the horizontal at which the projectile was launched = 30°
g = acceleration due to gravity = 9.8 m/s²
R = (30² sin 60°) ÷ 9.8
R = 78.53 m
So, Normally, the stone should travel a horizontal distance of 78.53 m. So, travelling a horizontal distance of 32 m (less than half of what the range should be without air resistance) means that, the motion of the stone was impeded, hence, option E is correct.
There must be a horizontal wind opposite the direction of the stone's motion, because ignoring air resistance when calculating the horizontal range would yield a value greater than 32 m.
Hope this Helps!!!
Answer: Wet barometer - The tool works by measuring atmospheric pressure to predict incoming weather. Since the glass is only filled halfway with water, the other half is exposed to the atmosphere. When the outdoor atmospheric pressure rises, the pressure in the glass decreases, and causes the water to move down the spout.
Dry barometer - A Torricellian barometer (sometimes called a mercury barometer) is an inverted (upside-down) glass tube standing in a bath of mercury. Air pressure pushes down on the surface of the mercury, making some rise up the tube. The greater the air pressure, the higher the mercury rises.
I hope this helps!
Answer:
Staples, Bestbuy, Maybe Homedepot
Explanation:
The correct answer among the choices above is option C. <span>Beryllium, Be, and chlorine, Cl, form a binary ionic compound with a one-to-two ratio of beryllium ions to chloride ions.</span> The chemical formula for the compound of beryllium and chlorine is BeCl2.
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
