Nuclear reaction you are literally splitting an atom and in a chemical reaction you are not
Answer:
0.661 s, 5.29 m
Explanation:
In the y direction:
Δy = 2.14 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.661 s
In the x direction:
v₀ = 8 m/s
a = 0 m/s²
t = 0.661 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²
Δx = 5.29 m
Round as needed.
Explanation:
Given Data
Total mass=93.5 kg
Rock mass=0.310 kg
Initially wagon speed=0.540 m/s
rock speed=16.5 m/s
To Find
The speed of the wagon
Solution
As the wagon rolls, momentum is given as
P=mv
where
m is mass
v is speed
put the values
P=93.5kg × 0.540 m/s
P =50.49 kg×m/s
Now we have to find the momentum of rock
momentum of rock = mv
momentum of rock = (0.310kg)×(16.5 m/s)
momentum of rock =5.115 kg×m/s
From the conservation of momentum we can find the wagons momentum So
wagon momentum=50.49 -5.115 = 45.375 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
Speed of wagon=45.375/(93.5-0.310)
Speed of wagon= 0.487 m/s
Throwing rock backward,
momentum of wagon = 50.49+5.115 = 55.605 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)
speed of wagon= 0.5967 m/s
Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27
Simora [160]
Data D has the largest range.
Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23
Therefore, Data D has the largest range.
