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jok3333 [9.3K]
2 years ago
12

6) A boat crosses a river at a constant engine speed of 2.0 m/s under pointed Directly west. The river runs directly south at 2.

5 m/s. If the river is 126 meters wide, how long does it take the boat to cross the river?
7) If the pilot of the boat doesn't correct for the current, how far downstream will he land on the opposite shore?
Physics
1 answer:
TEA [102]2 years ago
8 0

Answer:ligma ;)

Explanation:

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On a summer afternoon, the sand on the beach can get very hot. When you step on the sand in bare feet, you can burn yourself.
8_murik_8 [283]
<span>Heat from the Sun is transferred to the sand without direct contact. This heat is then transferred to your feet by direct contact.</span>
8 0
3 years ago
Read 2 more answers
With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons
Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0
3 years ago
What are the two applications of electron beam welding
Georgia [21]

First of all, there are not <u>just</u> two applications that are solely applicable to the electron beam welding process. There are MANY.

Please visit out website at the URL below and you can click the "View Application" button under each  listed Industry segment to view case studies of commonly EB welded applications.

https://www.ptreb.com/electron-beam-welding-applications

And for more general information on our welding process, we have an informational section you can peruse as well:

https://www.ptreb.com/electron-beam-welding-information

Good luck with your assignment-  we are glad to hear they are teaching about EBW in high school!!!

5 0
4 years ago
The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

           g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

6 0
2 years ago
On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel
Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}

|v_{0.2}|=11.99 m/s\approx 12 m/s

Velocity of 0.4 kg mass

v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}

|v_{0.4}|=12.99 m/s\approx 13 m/s

Thus total Kinetic energy =\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}

Kinetic energy=48.2 J

8 0
3 years ago
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