A car is moving at a initial velocity of 20 meters per second accelerates at a rate of 1.5 meters per second squared for 4 secon
ds. What is the finally velocity
1 answer:
Answered using calculus.
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s
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ANSWER
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.
A) Image position: -19.3 cm
B) Image height: 1.5 cm, upright
Explanation:
A)
In order to calculate the image position, we can use the lens equation:
where
p is the distance of the object from the lens
q is the distance of the image from the lens
f is the focal length
In this problem, we have:
p = 13 cm (object distance)
f = 40 cm (focal length, positive for a converging lens)
So the image distance is
The negative sign means that the image is virtual.
B)
In order to calculate the image height, we use the magnification equation:
where
y' is the image height
y is the object height
In this problem, we have:
y = 1.0 cm (object height)
p = 13 cm
q = -19.3 cm
Therefore, the image heigth is
And the positive sign means the image is upright.