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Fofino [41]
3 years ago
12

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

ers/second in 3 seconds. What’s the acceleration of the motorcycle over this time?
Physics
1 answer:
galben [10]3 years ago
5 0
3 meters per second
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Identify the following as
bazaltina [42]
3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy
6 0
3 years ago
hii! i need answers for both of these! this is due tomorrow and i want to get it done so i can finish all my other work, thank y
d1i1m1o1n [39]

Answer:

The mass and velocity for kinetic energy. Potential Energy: How high an object is and the mass in kilograms or it is the weight in and how high an object is. There are two formulas to calculate potential energy, but the one with grams is used more often.

Explanation:

Hope this helps!

5 0
3 years ago
In nuclear physics wht units are used to measure the radius of an atom ?
Alecsey [184]
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer  10^-15 m )
6 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
after a large snowstorm you shovel 2000 kilograms of snow off your side walk in 1 hour. you lift the shovel to an average height
solmaris [256]

Lifting a mass to a height, you give it gravitational potential energy of

       (mass) x (gravity) x (height)  joules.

To give it that much energy, that's how much work you do on it.

If 2,000 kg gets lifted to 1.25 meters off the ground, its potential energy is

         (2,000) x (9.8) x (1.25) = 24,500 joules.

If you do it in 1 hour (3,600 seconds), then the average power is

           (24,500 joules) / (3,600 seconds) = 6.8 watts.

None of these figures depends on whether the load gets lifted all at once,
or one shovel at a time, or one flake at a time.

But this certainly is NOT all the work you do.  When you get a shovelful
of snow 1.25 meters off the ground, you don't drop it and walk away, and
it doesn't just float there. You typically toss it, away from where it was laying
and over onto a pile in a place where you don't care if there's a pile of snow
there. In order to toss it, you give it some kinetic energy, so that it'll continue
to sail over to the pile when it leaves the shovel.  All of that kinetic energy
must also come from work that you do ... nobody else is going to take it
from you and toss it onto the pile.


8 0
3 years ago
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