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1 year ago

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Can you check this? Samantha swam upstream for some distance in one hour. She then swam downstream the same river for the same d

istance in only 12 minutes. If the river flows at 4 mph, how fast can Samantha swim in still water?

2 answers:

disa [49]1 year ago

6 0

Choose the most logical value for the variable to represent. Let x= Kelli's swimming speed in still water

ElenaW [278]1 year ago

3 0

So her speed in still water will be 6mph :)

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The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mm hg.

Schach [20]

The atmospheric P is greater than the P in the flask, since the Hg level is lacking down lower on the side open to the atmosphere.

43.4 cm x (10 mm / 1 cm) = 435 mm

the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the manometer. This means that 1 mmHg = 17.2 mm of manometer liquid.

435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm Hg

The pressure in the flask is 755 - 25.3 = 729.7 mmHg.

729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.

4 0

1 year ago

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

FrozenT [24]

Answer:

The total distance traveled is 28 inches.
The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector $\hat{i}$ pointing in the west direction, the ant start at position

$\vec{r}_0 = 16 \ inch \ \hat{i}$

Then, moves to

$\vec{r}_1 = 29 \ inch \ \hat{i}$

so, the distance traveled here is

$d_1 = |\vec{r}_1 - \vec{r}_0 | = | 29 \ inch \ \hat{i} - 16 \ inch \ \hat{i} |$

$d_1 = | 13 \ inch \ \hat{i} |$

$d_1 = 13 \ inch$

after this, the ant travels to

$\vec{r}_2 = 14 \ inch \ \hat{i}$

so, the distance traveled here is

$d_2 = |\vec{r}_2 - \vec{r}_1 | = | 14 \ inch \ \hat{i} - 29 \ inch \ \hat{i} |$

$d_2 = | - 15 \ inch \ \hat{i} |$

$d_2 = 15 \ inch$

The total distance traveled will be:

$d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch$

The displacement is the final position vector minus the initial position vector:

$\vec{D}=\vec{r}_2 - \vec{r}_1$

$\vec{D}= 14 \ inch \ \hat{i} - 16 \ inch \ \hat{i}$

$\vec{D}= - 2 \ inch \ \hat{i}$

This is 2 inches to the east.

6 0

2 years ago

The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig

Virty [35]

Answer:

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a<em> "net force"</em> which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. <u>So this explains number 2 & 3 answers.</u>

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). <u>So, this explains number 1.</u>

6 0

1 year ago

A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v

slava [35]

<u>Answer:</u>

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

<u>Explanation:</u>

Initial location of spaceship = (600 i - 400 j + 200 k)* $10^3km$ = (600 i - 400 j + 200 k)* $10^6m$

Initial velocity = 9500 i m/s

Acceleration = (40 i - 20 k) $10^3m/s^2$

Time = 35 minute = 35 * 60 = 2100 seconds

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Substituting

$s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m$

So final position = $((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m$

= $(708.15 i - 444.1 j + 200 k)*10^3km$

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

3 0

1 year ago

Read 2 more answers

The molecule arrangement of ice, water and steam are very different. _______ has the most ordered arrangement, ______ has the ne

marysya [2.9K]

Answer:

do you still need an answer

Explanation:

because if you are i could give it to you

3 0

2 years ago