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oee [108]
3 years ago
5

A person is going to take a trip from town eight to town D to do this they travel 60 KM in .45 hour And then stop in town before

lunch it takes. 15 hour in the journey continues to town see over .5 hour and covers a distance of 30 KM they had to stop and pick up an item at the store that took .15 hour they restart their trip it took 1.5 hours to get to town which is 90 came away from town see what was a numerical value for the average speed of the trip
Physics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

   v = 72 km / h

Explanation:

The definition of average speed is the distance traveled between the time interval

          v = Δx / Δt

let's find the distance traveled

       x = 60 + 30

       x = 90 km

time spent, all time must be included, travel time and when stopped

       t = 0.45 + 0.15 + 0.5+ 0.15

       t = 1.25 h

we substitute in the initial equation

      v = 90 / 1.25

       v = 72 km / h

in going from one city to the other

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A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

The height of the bullet at this time is calculated as follows;

y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft

Learn more about height of projectiles here: brainly.com/question/10008919

6 0
2 years ago
A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s
valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

m = 20kg

F = 50 Newtons (kg*m/sec^2)

acceleration, a, is provided for x seconds to increase the speed from 1 m/s to  3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2.  The cart increases speed by 2.5 m/s every second.  

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

7 0
2 years ago
Please help me on this!
aalyn [17]

Answer:

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Explanation:

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5 0
2 years ago
Road rage is an aggressive driving incident where the driver has:
WARRIOR [948]
I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)

5 0
3 years ago
A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found
kati45 [8]

Answer:

1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
  • Distance between the rod and the balloon, d = 0.0640 m.
  • Acceleration due to gravity, \rm g = 9.81\ m/s^2.

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

For the elecric force and the weight to be balanced,

\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

3 0
2 years ago
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