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storchak [24]
3 years ago
10

Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to t

he same point P. What is the ball's net displacement and the distance traveled?​
Physics
1 answer:
NikAS [45]3 years ago
5 0

Answer:

Net displacement = 0

Distance traveled = 2PQ <_up and down

Explanation:

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Which of these is not shared by bohrs model and the modern atomic model?
KonstantinChe [14]

Answer:

D. both have electrons that orbit the atomic nucleus in a simaler way

Explanation:

the bohrs model electrons orbit in a circular pattern were as the modern atomic model orbit in every which way

3 0
3 years ago
4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.
KIM [24]

Answer:

The K.E of the bowling ball right before it hits the ground, K.E = 2450 J            

Explanation:

Given data,

The mass of the bowling ball, m = 10 kg

The height of the building, h = 25 m

The total mechanical energy of the body is given by,

                                     E = P.E + K.E

At height 'h' the P.E is maximum and the K.E is zero,

According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'

Therefore, P.E at 'h'

                                  P.E = mgh

                                         = 10 x 9.8 x 25

                                         =  2450 J

Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J                                                                      

3 0
3 years ago
Which statement is true about an object that sinks in water?
bogdanovich [222]
It is about density. The less dense object will tend to float when put into the water. For example, oil is denser than water so the result of the mixture would make oil float. Sinking objects would only mean that it has an equal or greater mass than water.
7 0
4 years ago
Covalent bonds ___ conduct electricity well. <br><br> Always <br> Usually<br> Rarely<br> Never
mr_godi [17]
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4 0
3 years ago
Read 2 more answers
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
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