Answer:
a=7.384 m/s^2
Explanation:
let T be the tension in the string, m= mass
and a= acceleration
from the FBD in the attachment we can write
Tcos37°= mg
Tsin37° =ma
dividing both the equations we get
tan37° =a/g
therefore a=g×tan37°
a= 9.81×0.7535 = 7.384 m/s^2
the magnitude of the acceleration a of the train = 7.384 m/s^2
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = - 1.4 m
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The of the United States serves as the presiding officer of the Senate.
At present, is the president of the Senate.