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4 years ago

8

Select the game which is played on a 5' by 9' surface area

1 answer:

Lunna [17]4 years ago

6 0

Explanation:

There may be other choices, since the answer choices are not provided.

Table-tennis is played on a 5'x9' table by two or 4 players. The table is divided into two sections each 4.5'x5'. The sport is also called Ping-Pong.

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A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

antiseptic1488 [7]

Fnet=m*a
a= Fnet/m

a=(20N[E]+9.0N[E])/ 12.0kg

a=2.4167m/s^2

a ≈ 2.4m/s^2

8 0

4 years ago

Read 2 more answers

A client is receiving an IV solution of sodium chloride 0.9% (Normal Saline) 250 ml with amiodarone (Cordarone) 1 gram at 17 ml/

tester [92]

Answer:

1.1mg/min

Explanation:

We are given that

Volume of solution=250 ml

Mass of amiodarone=1 g

Infusion rate=17 ml/hr

We know that

1 g= 1000 mg

Ratio of 1000g:250 ml= $\frac{1000}{25}=4 mg/ml$

The concentration of solution=4mg/ml

Amiodarone infusing (mg/min)= $\frac{infusion \;rate(ml/hr)\times concentration}{60}$

Because 1 hr= 60 minute

Amiodarone infusing=1.1mg/m

Hence, 1.1 mg/min of amiodarone is infusing.

6 0

3 years ago

What is the thickness of the disk of the milky way?

Temka [501]

its average thickness is 10,000 light-years,

3 0

4 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

A wildlife researcher is tracking a flock of geese. The geese fly 4.0km due west, then turn toward the north by 40 degrees and f

BaLLatris [955]

Answer:

(a). The distance from the initial position is 6.68 km.

(b). The magnitude of their displacement is 7.05 km.

Explanation:

Given that,

Geese fly 4.0 km due west, then turn to north by 40° and fly another 3.5 km.

(a). We need to calculate the distance from the initial position

Using formula of distance

$D=AB+BC\cos\theta$

Put the value into the formula

$D=4.0+3.5\cos40$

$D=6.68\ km$

(b). We need to calculate the magnitude of their displacement

Using formula of displacement

$AC=\sqrt{CD^2+AD^2}$

$AC=\sqrt{(3.5\sin40)^2+(6.68)^2}$

$AC=7.05\ Km$

Hence, (a). The distance from the initial position is 6.68 km.

(b). The magnitude of their displacement is 7.05 km.

8 0

3 years ago

Read 2 more answers