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vagabundo [1.1K]
3 years ago
5

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

e to reach a pier on the river’s other side. To do so, the boat heads upstream at an angle θ = 32◦ from the direction to the boat’s pier. Find the ratio of vb to vr, where vr is defined above and vb is the boat’s speed with respect to the water
Physics
1 answer:
vova2212 [387]3 years ago
7 0

Answer: Vb is the vector  (-5.37m/s,  8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s,  8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

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Compute the torque about the origin of the gravitational force F--mgj acting on a particle of mass m located at 7-xî+ yj and sho
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Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}

\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)

[tex]\overrightarrow{\tau  }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

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3 years ago
A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.
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Answer:33

Explanation:

F = frequency

N =  Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V =  (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

8 0
3 years ago
3. A dog walks a distance of 100 ft. in 20 s of time. What is the dog's
natta225 [31]
The answer is A.
100/20=5
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3 years ago
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You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro
guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0
3 years ago
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A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.
Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0
3 years ago
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