Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Answer:
1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo
Explanation:
The given parameters are;
The mass of Christian and his bicycle = 54 kg
The mass of the cargo = 12 kg
1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity
∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N
2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N
Therefore, Christian has to exert more force on his bike after picking up the cargo.
Answer:
Explanation:
Work is defined as the scalar product of force and distance
W=F•d
Given that
F = 8.5i + -8.5j. +×-=-
F=8.5i-8.5j
d = 2.5i + cj
If the work in the practice is zero, then W=0
therefore,
W=F•ds
0=F•ds
0=(8.5i -8.5j)•(2.5i + cj)
Note that
i.i=j.j=k.k=1
i.j=j.i=k.i=i.k=j.k=k.j=0
So applying this
0=(8.5i -8.5j)•(2.5i + cj)
0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)
0=21.25-8.5c
Therefore,
8.5c=21.25
c=21.25/8.5
c=2.5
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