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3 years ago

14

While drag racing out of the parking lot I timed myself at a speed of

1 answer:

VladimirAG [237]3 years ago

3 0

Answer:

maybe ill be tracer

Explanation:

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DedPeter [7]

Answer:

A

Explanation:

A. The molecules that make up olive oil are longer than those that

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3 years ago

Read 2 more answers

A 15kg object is travelling 10m/s and hit a stationary 10kg object and stuck together. What is the final velocity of the objects

qaws [65]

Answer:

first object final velocity =2m/s

second = 12m/s

Explanation:

i hope this will help you,..,

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2 years ago

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was trav

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what was the speed limit, and it was 40 mph

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3 years ago

An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential differenc

Semenov [28]

Answer:

The kinetic energy is $KE = 5.67*10^{-18} \ J$

Explanation:

From the question we are told that

The potential difference is $\Delta V = 36 \ volts$

The potential energy of the end is mathematically represented as

$PEs = - q * \Delta V$

q is the charge on an electron with a constant value of $q = 1.60 *10^{-19} \ C$

substituting values

$PE = - 1.60*10^{-19} * 36$

$PE = - 5.67*10^{- 18} \ J$

Now from the law of energy conservation

The $PE_e = KEe$

Where $KE _e$ is the potential energy at the end

So

$KE = 5.67*10^{-18} \ J$

The negative sign is not includes because kinetic energy can not be negative

7 0

3 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

8 0

3 years ago