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3 years ago

a car moving at 30.0 m/s slows uniformly to a speed of 10.0 m/s in a time of 5.00 s determine the. acceleration of the car

Physics

1 answer:

netineya [11]3 years ago

7 0

Answer: $-4m/s^{2}$

Explanation:

Acceleration $a$ is defined as the variation of Velocity $\Delta V$ in time $\Delta t$ :

$a=\frac{\Delta V}{\Delta t}$

In this case $\Delta V=10m/s - 30m/s$ and $\Delta t=5 s$ .

Therefore:

$a=\frac{10m/s - 30m/s}{5 s}$

$a=\frac{-20m/s}{5 s}$

$a=-4 m/s^{2}$ Note the negative sign indicates the car is slowing

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In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

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Answer:

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Explanation:

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Change in current

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A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

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Answer:

a) see attached, a = g sin θ

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))

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