The rock's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = 45 m + (7.2 m/s) <em>t</em> - 1/2 <em>g</em> <em>t </em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Set <em>y</em> = 0 and solve for <em>t</em> :
0 = 45 m + (7.2 m/s) <em>t</em> - 1/2 <em>g</em> <em>t </em>² → <em>t</em> ≈ 3.9 s
The answer is the third choice, "the brightness of the beam of light increases"
According to Einstein’s theory, an increase in the number of photons (per unit are) affects a beam of light by causing a higher intensity. <span> Sometimes, the term "brightness" is used when referring to the </span>intensity<span> of a color, a</span><span>lthough there are instances where this can be a misleading term when we try to describe </span><span>intensity</span>
Answer:
It spreads out into multiple shadow regions
Explanation:
- when the waves meet the barrier than light is diffracted as When this occurs, the wave bends around the corners of the barrier or passes through the opening of the wedge, which acts as a barrier, forming several patterns with the hole shape of the wedge.
- and The main condition for this phenomenon to occur is that the magnitude of the barrier must be equal to the magnitude of the wavelength.
- when sunlight as an electromagnetic waves is passes by key hole of gate, then this light will break and form many keyhole-shaped shadow fields.
Answer:
d = 136.7 ft
Explanation:
Because the truck move with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (ft)
v₀: initial speed in ft/s
vf: final speed in ft/s
a: acceleration in ft/s²
Data
v₀ = 44.0 mi/h
1milla = 5280 ft
1h = 3600 s
v₀ = 44*(5280 ft) / (3600 s) = 64.5 ft/s
vf = 0
d = 47.0 ft
Calculation of the acceleration of the truck
We replace data in the formula (1) :
vf²=v₀²+2*a*d
0 = (64.5)²+2*a*(47)
-(64.5)² = (94)*a
a = -(64.5)² / 94
a = - 44.26 ft/s²
The acceleration (a) it's negative (-) because the truck is braking
Calculation of the minimum stopping distance of the truck to v₀ = 75.0 mi/h
v₀ = 75 mi/h = 75* (5280 ft) / (3600s) = 110 ft/s
We replace v₀ = 110 ft/s and a = - 44.26 ft/s² in the formula (1):
vf²=v₀²+2*a*d
0 = (110)²+2*(-44.26)*d
88.52*d = (110)²
d = (110)² / (88.52)
d = 136.7 ft