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2 years ago

8

A block of wood has density 0.500 g/cm3 and mass 2 000 g. It floats in a container of oil (the oil's density is 0.750 g/cm3). Wh

at volume of oil does the wood displace

1 answer:

Gennadij [26K]2 years ago

6 0

Answer:

2,666.67cm^3

Explanation:

All we need to do in this problem is divide the mass of the wooden block to the oil's density.

2000/0.750 ≈ 2,666.67cm^3

Best of Luck!

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A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

Order the layers, with the oldest at the bottom and most recent at the top.

givi [52]

Answer:

adbce i think

Explanation:

7 0

3 years ago

This is for physics, i need help fast!!

jarptica [38.1K]

Speed = distance / time
S= 40 000m / 5400s
S= 7.41m/s

7 0

3 years ago

The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be

eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0

2 years ago