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Natali [406]
2 years ago
9

A boat is heading across a river at a velocity of 25 mph. The river is flowing downstream at 10 mph. What is the actual velocity

of the boat?
Physics
1 answer:
AnnyKZ [126]2 years ago
8 0
Principle: Pythagorean theorem

Given:
Let a = boat velocity = 25 miles per hour
      b = river velocity = 10 miles per hour

Required: c = actual velocity

Solution:
     
     c^2 = a^2 + b^2
     c^2 = (25)^2 + (10)^2
     c^2 = 725
     (c^2)^0.5 = (725)^0.5
      c = 26.93 miles per hour

<em>ANSWER: Actual boat velocity is 26.93 miles per hour</em>
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A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle
alexira [117]
<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
  • According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
  • Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

       = 0.3759

r = Sin^-1 0.3759

   = 22.08

   <u>≈ 22°</u>

3 0
2 years ago
A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height
Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

v_x = 23.2 m/s

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

v_y(t)= v_{y0} -gt (1)

where

v_{y0}=0 is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s

Substituting t into (1) we find the final vertical velocity

v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}

6 0
2 years ago
Sarah is 14 years old and skips school twice a week without any written explanation. What can she be charged with?
harina [27]

Answer:

Too many people are unaware or indifferent to that.” Fines can cost up to $500 per truancy, due within 30 days unless a judge gives an extension. For many students and families, it's another debt they can't pay. And if fines aren't paid, they can convert into an arrest warrant when a student turns 17.

Explanation:

5 0
2 years ago
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
2 years ago
In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their ki
myrzilka [38]

Answer:

a) 2 x10^7 eV

b) 2 x10^4 keV

c) 20 MeV

d) 0.02 Gev

e) 3.2 x 10^-12J

Explanation:

The potential difference = 20 x 10^6 V

The charge on the proton = 1.6 x10^-19

The work done to move the proton will be basically the proton will acquire if it accelerates.

Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19

                                                 =3.2 x 10^-12J or 2 x10^7 eV

2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev

5 0
2 years ago
Read 2 more answers
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