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Leto [7]
3 years ago
5

Help, im not sure how to do this at all

Physics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

60 m

Explanation:

The boat has two separate motions:

1- A motion due north, with constant speed of 10 m/s

1- A motion due east, due to the current, at speed of 2 m/s

We know that the river is 300 m wide, so we can consider first motion 1) to find how much does it take for the boat to cross the river:

t=\frac{d}{v}=\frac{300 m}{10 m/s}=30s

Now we can find how far downstream the boat moved by calculating the distance that the boat covered moving east during this time interval:

d'=vt=(2 m/s)(30 s)=60 m

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Tema [17]
The answer is B............
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3 years ago
Can someone help me figure out how to find tension in a rope? The question is:
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You are welcome.......

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3 years ago
2kg of water at 30 C is poured into a 1kg copper beaker at 20 C.
GREYUIT [131]

The temperature of the water and copper beaker be together is 29.6⁰C.

<h3>What is the equilibrium temperature of both substance?</h3>

The final temperature or equilibrium temperature of the water and copper beaker is calculated by applying the principle of conservation of energy.

Heat lost by the water = Heat gained by the copper beaker

mcΔθ (water) = mcΔθ (copper)

where;

  • m is mass
  • c is specific heat capacity
  • Δθ is change in temperature

m₁c₁(T₁ - T) = m₂c₂(T - T₂)

where;

  • T₁ is the initial temperature of water
  • T₂ is the initial temperature of copper beaker
  • T is the equilibrium temperature

Specific heat capacity of copper, c₂ = 389 J/kgK

Specific heat capacity of water , c₁ = 4200 J/kgK

(2)(4200)(30 - T) = (1)(389)(T - 20)

252,000 - 8400T = 389T - 7780

259,780 = 8789T

T = 259,780 /8789

T = 29.6⁰C

Learn more about equilibrium temperature here: brainly.com/question/8925446

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8 0
2 years ago
An eagle is flying horizontally at a speed of 3.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.10 m
Alinara [238K]

Answer:

10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)

Explanation:

The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

We use the first equation of motion for a free-falling body to obtain v as follows;

v = u + gt....................(1)

where g is acceleration due to gravity taken as 9.8m/s/s

It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

h=ut+gt^2/2.................(2)

Given; h = 6.10m.

since u = 0 for the vertical motion;  equation (2) can be written as follows;

h=\frac{1}{2}gt^2............(3)

substituting;

6.1=\frac{1}{2}*9.8*t^2\\6.1=4.9t^2\\hence\\t^2=6.1/4.9\\t^2=1.24\\t=\sqrt{1.24}=1.12s

Putting this value of t in equation (1) we obtain the following;

v = 0 + 9.8*1.12

v = 10.93m/s

5 0
3 years ago
Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00
lions [1.4K]

Answer:

b.  \Delta KE = 390 eV

Explanation:

As we know that the electric field due to infinite line charge is given as

E =\frac{\lambda}{2\pi \epsilon_0 r}

here we can find potential difference between two points using the relation

\Delta V = \int E.dr

now we have

\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr

now we have

\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})

now plug in all values in it

\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})

\Delta V = 216ln6 = 387 V

now we know by energy conservation

\Delta KE = q\Delta V

\Delta KE = (e)(387V) = 387 eV

3 0
3 years ago
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