Answer:
90m
Explanation:
Distance is not a vector but a scalar quantity. depicted by a quantity only.
hence:
Distance is equal to 30+40+20 = 90 m
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Answer:
Explanation:
Given that,
A vector A has x component to be 2.7cm and y component to be 2.25cm
Then,
A = 2.7•i + 2.25•j
A vector B has x component of 0.30cm and y component of 1.75cm
B = 0.3•i + 1.75•j
So, we want to find A+B
Addition of vectors
Generally
(a•i + b•j) + (c•i + d•j) = (a+c)•i +(b+d)•j
Vectors are added component wise.
So,
A + B = (2.7•i + 2.25•j) + (0.3•i + 1.75•j)
A + B = (2.7 + 0.3)•i + (2.25 + 1.75)•j
A + B = 3•i + 4•j
We can also find it magnitude and direction
Generally,
A = a•i + b•j
|A| = √(a²+b²)
<A = arctan(b/a)
So,
|A+B| = √(3²+4²) = √9+16 = √25
|A+B| = 5
And it's direction
< = arctan(y/x)
< = arctan(4/3)
< = 53.13°
This movement of the coin relative to the knee is not under an inertial frame, because knee is accelerated relative to the coin.
a) coin acceleration relative to the ground: g = 9.81 m/s^2. This is the acceleration due to the atraction of the earth, given that the coin is not tied to the car or to the knee.
b) coin acceleration relative to the knee: acceleration of the coin relative to the ground - acceleration of the knee relative to the ground = - g - (-1.24g) = 0.24 g = 0.24 * 9.81 m^2^2 = 2.354 m/s^2
c) time to the coin move 2.2 m upward, relative to the car
d = a*t^2 / 2 , where a is the acceleration relative to the car (same of the knee)
=> t = √(2d/a) = √(2*2.2m/2.354 m/s^2) = 1.37 s
d) actual force on the coin
Use the acceleration relative to the ground, a=g = 9.81 m/s^2
F = m*a = 0.567 g * 10^-3 kg/g * 9.81 m/s^2 = 5.56*10^-3 N
e) apparent force
Use apparent acceleration, a = 2.354 m/s^2
F = m*a = 0.567*10^-3 kg * 2.354 m/s^2 = 1.33*10^-3 N
Answer:
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