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jarptica [38.1K]
3 years ago
7

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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C, They change their shapes depending on their containers

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Draw the Lewis dot Structure for the molecule C2H6​
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This means that the Lewis dot structure for C2H6 must account for 14 valence electrons, either through bonding between atoms, or through lone pairs. So, the two C atoms are placed in the center of the molecule.

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If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit
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Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

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4 0
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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:

2.083 V.

Explanation:

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Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

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requirement of V = (500 / 6 )  x 25 =   2083.33 mV = 2.083 V.

7 0
3 years ago
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
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

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

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              v = √(2gL (1-cos θ))

4 0
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