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Ket [755]
1 year ago
14

A river has a steady speed of vs. A student swims upstream a distance d and back to the starting point. (a) If the student can s

wim at a speed of v in still water, how much time tup does it take the student to swim upstream a distance d? Express the answer in terms of d, v, and vs. tup = Incorrect: Your answer is incorrect. (b) Using the same variables, how much time tdown does it takes to swim back downstream to the starting point? tdown = Incorrect: Your answer is incorrect. Your answer cannot be understood or graded. More Information
Physics
1 answer:
andre [41]1 year ago
6 0

he speed of the student relative to shore is

v_ up = v- vs

v _down = v+ vs

The time required to travel distance d upstream
is

t_up = d/ v_up = d/ v- vs

(2)

The time required to swim the same distance d downstream is

t_down = d/ v_down = d/ v+ vs

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Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
2 years ago
An artificial satellite in a low orbit circles the earth every 98 minutes. what is its angular speed in rad/s?
postnew [5]
To finish one orbit it will take 98 x 60 seconds. So; <span>(2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec. </span><span>
</span>
8 0
2 years ago
The burj Khalifa in Dubai is the worlds tallest building. It rises to an amazing 828M above the ground and if you were to get to
antoniya [11.8K]
I don't understand what you mean by "depth" of the steps.  The flat part of the step has a front-to-back dimension, and the 'riser' has a height.  I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy.  And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground.  So something is definitely fishy about the steps.

Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.

In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters.  The work [s]he does is (force) x (distance) = <em>662,400 joules. </em>
6 0
2 years ago
A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
liq [111]

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

5 0
1 year ago
A 900 kg steel beam is supported by the two ropes shown in (Figure 1) . Calculate the tension in the rope.
Rzqust [24]
Let T1 and T2 be tension in ropes1 and 2 respectively. 
<span>since system is stationary (equilibrium), considering both ropes + beam as a system </span>

<span>for horizontal equilibrium (no movement in that direction, so resultant force must be zero horizontally) </span>
<span>T1sin(20) = T2sin(30) </span>
<span>=> T1 = T2sin(30) / sin(20) </span>

<span>for vertical equilibrium, (no movement in this direction, so resultant force must be zero vertically) </span>
<span>T1cos(20) + T2cos(30) = mg </span>

<span>m = 900kg, substituting for T1 </span>
<span>T2sin(30)*cos(20)/sin(20) + T2cos(30) = 900g </span>
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<span>T2 = 3788.65N </span>
<span>so T1 from (1) </span>
<span>T1 = 5535.21N</span>
8 0
2 years ago
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