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Anastaziya [24]
3 years ago
13

Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

of the resulting velocity?
a)

14 m/s South


b)

14 m/s North


c)

10 m/s North


d)

2 m/s North
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

2 /s north

Explanation:

Given that,

Velocity due North is 8 m/s and due south is 6 m/s

We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

v=8+(-6)\\\\v=2\ m/s

It is positive. So, it is in North direction.

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When something is hit harder how does the transverse wave change?
Flura [38]
When something is hit harder just like when sound is turned up the waves become higher and more frequent like a zig zag more so then wavy.
5 0
2 years ago
A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

 v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}

 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
Which is normally greater, the energy in ordinary sound or the energy in ordinary light? how does the speed of sound compare to
Thepotemich [5.8K]

The energy in ordinary light is greater than the energy in ordinary sound

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Energy is a physical system's ability to perform labor. The capital letter E is a typical sign for energy. The joule, denoted by the letter J, is the common unit. The energy produced by one newton's (1 N) worth of force acting over one meter's (1 m) worth of displacement is measured in joules (1 J). Because it is a fundamental human requirement, energy plays a significant role in our daily lives.

To learn more about Energy please visit-
brainly.com/question/1932868
#SPJ4

8 0
2 years ago
A cold front moves through and the temperature drops by 20 degrees. in which temperature scale would this 20 degree change repre
nalin [4]
In the Celsius scale each degree is one part of 100 degrees. This is because in this scale the difference between boiling and freezing temperatures of water is 100 ° - 0 ° = 100 °, so one degree Celsius is one part of 100.

In the Farenheit scale, each degree is one part of 180 degrees. This is because in this scale the difference between the boiling and freezind temperatures are 212 ° - 32 ° = 180°, so one degree Farenheti is one part of 180.

That means that 1 °C is a larger amount than 1 °C, so 20°C is a larger amount than 20°F.

Conclusion: 20 degree change represents a larger change in Celsius scale.

4 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
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