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3 years ago

Turning up the Heat

Physics

1 answer:

ella [17]3 years ago

3 0

Answer:

harem jath

Explanation: t=i dont know anything about how to speak english thats the answer for your question

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A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago

This is for 6.02 in comprehensive science for flvs

Katen [24]

Answer:

Please provide an image to help clarify

Explanation:

thanks :)

4 0

3 years ago

The bending of a wave as it enters a new medium at an angle is called _____.

Andru [333]

Answer:

C). Refraction

Explanation:

When light enters from one medium to another medium then it changes it's direction from its initial direction

the phenomenon of changing the direction due to change in the medium is known as Refraction of light

So here we know that when light strikes an interface which separates two medium then due to change in the medium we will have

$\mu_1 sin\theta_1 = \mu_2 sin\theta_2$

so here we have

$\theta_1, \theta_2$ = two angles of the light in two mediums

$\mu_1 , \mu_2$ = two refractive index of the two mediums

So correct answer will be

C). Refraction

4 0

3 years ago

You are riding your bike at 9 km/h twn minutes later your speed is 6 km/h

gtnhenbr [62]

De-celleration because speed is lower as time goes on

5 0

3 years ago

Read 2 more answers

Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i

fiasKO [112]

Answer:

The algebraic equation is:

$F_{v} =\frac{m_{b}v_{b}^{2} }{R} -m_{b} g$

Explanation:

Given information:

mb = book's mass

vb = tangential speed

R = radius of the path

Question: Derive an algebraic equation for the vertical force, Fv = ?

To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:

$F_{v} =\frac{m_{b}v_{b}^{2} }{R} -m_{b} g$

The variables were defined above and g is the gravity.

4 0

4 years ago