Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
Answer:
The correct answer is t = 0.92s
Explanation:
Initial velocity v0 = 3.0 m/s
Displacement Δy = ?
Acceleration a = -9.8m/s2
Final velocity v = -6.0m/s
Time t=? Target unknown
We can use the kinematic formula missing Δy to solve for the target unknown t:
V=v0+at
We can rearrange the equation to solve to t:
V-v0=at
t= v-v0/a
Substituting the known value into the kinematic formula gives:
t= (-6.0m/s)-(3.0m/s)
————————————
-9.8m/s2
= -9m/s
—————-
-9.8m/s2
=0.92s
Answer:
violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm,
The infrared (IR) range wavelengths greater than 700 nm
Explanation:
The spectral region near the visible is what humans can see with our eyes.
The range of the ultra violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm, this radiation is responsible for tanning and skin burns.
The infrared (IR) range is a very wide range that begins at wavelengths greater than 700 nm and continues up to approximately 25,000 nm (1012 Hz).
