A force of N is applied to an object. How much work is done, in Joules, moving the object from x=1 to x=4 meters?
1.
2. 51 J
3.
4.
A car has a certain displacement between 0 seconds and 2 seconds. If we defined its velocity as v(t)=6t-5, then the displacement in meters is:
1. 1 m
2. 2 m
3. 3 m
4. 4 m
The relation between time t and distance x is , where a and b are constants, the acceleration is
(1)
(2)
(3)
(4)
Given velocity v(t) = . Assume s(t) is measured in meters and t is measured in seconds. If s(0) = 0, the position s(4) at t = 4s is:
1. | \(30\) | 2. | \(31\) |
3. | \(32\) | 4. | \(33\) |
The current through a wire depends on time as \(i = (2+3t)~\text{A}\).
The charge that crosses through the wire in \(10\) seconds is: \(\left(\text{Instantaneous current,}~i= \frac{dq}{dt} \right)\)
1. \(150~\text{C}\)
2. \(160~\text{C}\)
3. \(170~\text{C}\)
4. None of there
The area of a blot of ink, \(A\), is growing such that after \(t\) seconds, \(A=\left(3t^2+\frac{t}{5}+7\right)\text{m}^2\). Then the rate of increase in the area at \(t = 5~\text{s}\) will be:
1. \(30.1~\text{m}^2/\text{s}\)
2. \(30.2~\text{m}^2/\text{s}\)
3. \(30.3~\text{m}^2/\text{s}\)
4. \(30.4~\text{m}^2/\text{s}\)
A particle starts rotating from rest and its angular displacement is given by \(\theta = \frac{t^2}{40}+\frac{t}{5}\). Then, the angular velocity \(\omega = \frac{d\theta}{dt}\) at the end of \(10~\text{s}\) will be:
1. \(0.7\)
2. \(0.6\)
3. \(0.5\)
4. \(0\)
, where C is a constant, can be expressed as:
1.
2.
3.
4.
If the force on an object as a function of displacement is \(F \left(x\right) = 3 x^{2} + x\), what is work as a function of displacement \(w(x)\): \(\left(w= \int f\cdot dx\right)\) Assume \(w(0)= 0\) and force is in the direction of the object's motion.
1. \(\frac{3 x^{3}}{2} + x^{2}\)
2. \(x^{3} + \frac{x^{2}}{2}\)
3. \(6x+1\)
4. \(3 x^{2} + x\)