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strojnjashka [21]
3 years ago
10

Here's a great everyday use of the physics described in Think about what subsequently happens to the ketchup, which is initially

at rest, and use Newtons first law to explain why this technique is so successful this chapter. If you are trying to get keichup out of the bottle, the best way to do it is to turn the bottle upside down and give the bottie a sharp upward smack, forcing the bottle rapidly upward. (Figure 1 The weight of the ketchup will keep it from moving if you give the bottle a sharp upward smack Figure The static friction of the ketchup wil keep it from moving if it isnt too tightly adhered to the sides of the moving bottle of1 The inertia of the ketchup will keep it from moving if it isn't too tightly adhered to the sides of the moving bottle. The kinetic friction of the ketchup will keep it from moving if it isn't too tightly adhered to the sides of the moving bottle
Physics
2 answers:
balandron [24]3 years ago
6 0

Answer:

Explanation:

The inertia of the ketchup will keep it from moving if it isn't too tightly adhered to the sides of the moving bottle.

Andrei [34K]3 years ago
5 0

Answer: "The static friction of the ketchup wil keep it from moving if it isnt too tightly adhered to the sides of the moving bottle"

Explanation:

We want to see why is that the ketchup does not move of the bottom of the bottle when we want to use it.

The options are:

Because of the weight.

Because of the static friction.

Because of the inertia.

Because of the kinetic friction.

First, we can discard weight and inertia, because these two things actually help to get the ketchup out of the bottle.

The remaining options are static friction and kinetic friction:

If the kinetic friction has an effect, it means that the ketchup inside the bottle is moving, so this option can also be discarded.

Then the correct option is static friction, which the ketchup does against the walls of the bottle and keeps it in place.

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The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig
miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0
3 years ago
A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance
Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, m_b=11\ g=0.011\ kg

Mass of the pendulum, m_p=19\ kg

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

m_bv=(m_p+m_b)V

Put the value of V from equation (1) in above equation as :

v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s

So, the bullet's initial speed is 243.21 m/s.

7 0
3 years ago
What is an Atwood Machine?
Lady_Fox [76]
The best answer is letter (A) a double pulley system. Atwood Machine is normally used as a measurement in balancing to object to verify the mechanical law of motion with constant acceleration.
6 0
3 years ago
A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during
Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

  • From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
  • From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

  • The speed increases from 0 m/s to 1341 m/s.
  • The time elpased is 2.0 min.
  • 1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s}  = 11 m/s^{2}

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

3 0
2 years ago
What is the term used to label the energy levels of electrons?
mel-nik [20]
Electron volts...........
4 0
3 years ago
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