12. Answer the following questions using the velocity and time graph above.

A ball is dropped from rest and falls to the floor. The initial gravitational potential energy of the ball-Earth-floor system is

Cloud [144]

Answer:

The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

Explanation:

It is given that,

Initial gravitational potential energy of the ball-Earth-floor system is 10 J.

The ball then bounces back up to a height where the gravitational potential energy is 7 J.

Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.

Initial mechanical energy is 10 + 0 = 10 J

Mechanical energy just before the collision is 0 + 10 = 10 J

Final mechanical energy, 7 + 0 = 7 J

Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.

4 0

3 years ago

A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge

adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows $P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}$

thus where the pipe is horizontal we have

z₁ = z₂ hence the above equattion becomes

$P_{1} +\frac{1}{2}\rho v^{2} _{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2}$

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

5 0

4 years ago

The beginning of a river is called the<br> mouth<br> source<br> watershed<br> tributary

aalyn [17]

Answer:

head of water or mouth

Explanation:

<h3>hope this helps</h3>

8 0

4 years ago

Suppose you have made the following hypothesis: Sharks are most common near coral reefs, because there are more fish there to ea

Brilliant_brown [7]

I would say none of the options. This evidence does not support the hypothesis, but it doesn't contradict it, however it is related to the hypothesis. I guess what I'm trying to say is that the evidence isn't sufficient enough to make any definitive comments about the hypothesis. I don't think that you can just decide on whether to accept or reject your hypothesis based on observation alone and moreover, an observation that was made once. You need to make many observations, at certain points every day, in the same area of reef and the same area of open sea for a certain amount of time to gain a good amount of data (you could split up areas of reef and open sea on a particular coast into square meters or what ever unit you want and dedicate 3 days to each area you've split up) then you can perform a statistical test that suits the model of your data. I hope this helps in some way and I'm sorry it's so long. I couldn't think of a shorter way to say this.

3 0

4 years ago

Read 2 more answers

Need help ASAP..please help

jonny [76]

Answer:

option 3

Explanation:

can i get brainliest

4 0

3 years ago