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Thepotemich [5.8K]
3 years ago
7

Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 2 m. What is your acceleration?

Physics
1 answer:
likoan [24]3 years ago
7 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s

time for the change = 2 minutes = 120 seconds

Acceleration = (8 m/s) / (120 seconds)

Acceleration = 0.067 m/s²

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The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?
Harrizon [31]

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

5 0
2 years ago
A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin
Katarina [22]
I’m not sure look it up in the internet!
4 0
2 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

The angle is because the friction force opposes the movement

          Δk =K_{f} –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

8 0
3 years ago
In SI units, what is the magnitude the net force acting on a 1,152 kg car that accelerates uniformly along a straight line from
geniusboy [140]

Answer:

Fnet = 14515.2 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force.
  • Fapp is the applied force.
  • Fg is the force due to gravitation.

Given the following data;

Mass = 1,152 kg

Initial velocity, u = 3m/s

Final velocity, v = 17m/s

Time, t = 5 seconds

To find the magnitude of the net force;

First of all, we would determine the acceleration of the car.

Acceleration = (v-u)/t

Acceleration = (17 - 3)/5

Acceleration = 14/5

Acceleration = 2.8m/s

To find the applied force;

Fapp = mass * acceleration

Fapp = 1,152 * 2.8

Fapp = 3225.6 N

Next, we would find the force exerted on the car due to gravity.

Fg = mass * acceleration due to gravity

We know that acceleration due to gravity is equal to 9.8m/s²

Fg = 1152 * 9.8

Fg = 11289.6N

Substituting the values into the equation, we have;

Fnet = 3225.6 + 11289.6

Fnet = 14515.2 Newton

7 0
3 years ago
Does exists the friction force in space?If yes,tell an full example
VashaNatasha [74]
Yes, friction does exist in space. Friction has nothing to do with the earth's atmosphere. It exists everywhere in the universe. <span />
5 0
3 years ago
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