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Otrada [13]
4 years ago
9

A river flows at a rate of 2 km divided by h. A patrol boat travels 54 km upriver and returns in a total time of 9 hr. What is t

he speed of the boat in still​ water?
Physics
1 answer:
dolphi86 [110]4 years ago
4 0

Answer:

12.32 km/h

Explanation:

V_r=Velocity of river = 2 km/h

V_b=Velocity of boat

V_b-V_r = Speed of boat going against river

V_b+V_r = Speed of boat going along river

Distance to travel = 54 km

Total time taken = 9 hours

So,

\frac{54}{V_b-V_r}+\frac{54}{V_b+V_r}=9\\\Rightarrow \frac{54(V_b+V_r+V_b-V_r)}{V_b^2-V_r^2}=9\\\Rightarrow \frac{54(2V_b)}{V_b^2-V_r^2}=9\\\Rightarrow \frac{54(2V_b)}{9}=V_b^2-V_r^2\\\Rightarrow 12V_b=V_b^2-V_r^2\\\Rightarrow V_b^2-V_r^2-12V_b=0\\\Rightarrow V_b^2-12V_b-4=0

Solving this quadratic equation we get,

V_b=\frac{12\pm \sqrt{144+16}}{2}=12.32\ or -0.32

So, velocity of boat in still water is 12.32 km/h

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At rest, hydrogen has a spectral line at 116 nm. if this line is observed at 107 nm for the star sirius, how fast is sirius movi
Citrus2011 [14]

2.3275862×10¹²km/s fast is sirius moving in km/s.

<h3>Briefing:</h3>

Hydrogen has a spectral line at = 116nm=116×10⁻⁹m

Line is observed at = 107 nm=107×10⁻⁹m

Now, from the Hubble's law

V=(\Delta \lambda / \lambda)×C

Where,

v is the velocity

Δλ = Change in wavelength = 116 - 107= 9nm=9×10⁻⁹m

λ = Actual wavelength=116nm=116×10⁻⁹m

C is the speed of the light=3×10⁸m/s

on substituting the respective values, we get

V=(9/116)×3×10⁸=23275862.069×10⁵m/s

V=2.3275862×10¹²km/s.

<h3>What is the wavelength?</h3>

A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. Typically, in wireless systems, this length is specified in meters (m), centimeters (cm), or millimeters (mm).

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4 0
2 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
4 years ago
A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/
Artyom0805 [142]

Answer:

I=182.97\ kg-m^2

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, \tau=860\ N

Angular acceleration of the shell, \alpha =4.7\ m/s^2

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

\tau=I\alpha

I is the rotational inertia of the shell

I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2

So, the rotational inertia of the shell is 182.97\ kg-m^2.

7 0
3 years ago
How much is the velocity of a body when it travels 600m in 5 minutes?​
maks197457 [2]

answer

so unit of velocity is m/s

displacement=600m

5minutes should be converted to seconds

5×60=300 seconds

so,

velocity= displacement÷time

= 600m ÷300s

=2m/s or 2ms^-1

4 0
2 years ago
Read 2 more answers
The speed of the 2 kg cart after 5 seconds is ? cm/s.
motikmotik
The answer to this question is 1cm/s
3 0
3 years ago
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