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murzikaleks [220]
2 years ago
12

You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8

.5 seconds to stop?
Physics
1 answer:
Arlecino [84]2 years ago
6 0

You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

<u>Given the following data:</u>

  • Initial velocity, U = 31.3 m/s  
  • Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

  • S is the distance travelled.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the parameters into the formula, we have;

S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025

<em>Distance, S</em><em> = </em><em>620.075 meters.</em>

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

Read more: brainly.com/question/8898885

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Just need help with 1 and 2 please :D i’m having a bit of trouble :/
dexar [7]
1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
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D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
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In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

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