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KIM [24]
3 years ago
9

Walk in a straight line. Now stop. Did you accelerate? Explain

Physics
2 answers:
sasho [114]3 years ago
7 0
YES. If you where not moving and then suddenly started moving forward to walk in the straight line you did indeed accelerate at one point in your journey. And then when you stopped you where decelerating
Brums [2.3K]3 years ago
4 0

Answer;

It's about acceleration, right?

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The position of a 60 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4
mrs_skeptik [129]

The position of an oscillating mass is given by:

x(t)=A cos (\omega t)

where A is the amplitude of the oscillation, \omega the angular frequency and t the time.

The velocity of the oscillating mass can be found by calculating the derivative of the position:

v(t)=x'(t)=-\omega A sin (\omega t)

In this problem, A=2.0 cm and \omega=10 rad/s, so if we substitute these data and t=0.4 s we can find the velocity at t=0.4 s:

v(t)=-(10 rad/s)(2.0 cm) sin ((10 rad/s)(0.4 s))=-13.07 cm/s=-0.13 m/s

3 0
3 years ago
A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF.
Aleksandr-060686 [28]
Answer:
Q = 8 μC

Explanation:
The relation between voltage, capacitance and charge can be expressed using the following rule:
Q = C * V
where:
Q is the amount of charge that we want to calculate
C is the capacitance = 4 * 10⁻⁶ F
V is the voltage applied = 2 V

Substitute with the givens in the above equation to get the amount of charge as follows:
Q = C * V
Q= 4 * 10⁻⁶ * 2
Q = 8 * 10⁻⁶ Coulumb
Q = 8 μC

Hope this helps :)
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