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3 years ago

13

PLZ HELP ME!!!!!!!!! WILL GIVE BRAINLY!!!!!!!! Which photo represents a waxing gibbous moon?

1 answer:

yarga [219]3 years ago

6 0

A phase of moon with more than half sunlit portion visible on the right

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Maria designs a test to see if lemon trees that receive more water produce larger lemons.

denis-greek [22]

Do you know the answer

7 0

3 years ago

A ball is connected to a light spring suspended vertically. When pulled downward from its equilibrium position and released, the

babunello [35]

Answer:

The forms of energy involved are

1. Kinetic energy

2. Potential energy

Explanation:

The system consists of a ball initially at rest. The ball is pulled down from its equilibrium position (this builds up its potential energy) and then released. The released ball oscillates due to a continuous transition between kinetic and potential energy.

5 0

3 years ago

Which of the following are results of the force of gravity?

docker41 [41]

Hello,

Here is your answer:

The proper answer for this question is option B "When released,a book falls to the ground". That's because of gravity the book will hit the ground!

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps.

7 0

3 years ago

Read 2 more answers

Please i need help! ill give brainliest toooo!

Igoryamba

Answer:

b

Explanation:

7 0

3 years ago

Read 2 more answers

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago