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ELEN [110]
3 years ago
13

A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc

al ocean current is 1.50 m/s in a direction 40.0º north of east and changes the ship's intended motion. What is the velocity of the ship relative to the Earth?
Physics
1 answer:
sattari [20]3 years ago
5 0

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

\overrightarrow{v}_{s,w}=6.5 \widehat{j}

velocity of water with respect to earth = 1.5 m/s at 40° north of east

\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}

\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j}  \right )

\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}

The magnitude of the velocity of ship relative to earth is \sqrt{1.15^{2}+5.54^{2}} = 5.66 m/s

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Two 2.0 cm by 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires of the terminals of a 9.0 V battery.
ale4655 [162]

Answer:

Explanation:

Area of electrodes, A = 2 cm x 2 cm = 4 cm²

Separation between electrodes, d = 1 mm

Voltage, V = 9 V

(a)

Let C is the capacitance between the electrodes

C = \frac{\epsilon _{0}A}{d}

C = \frac{8.854\times 10^{-12}\times 4\times 10^{-4}}{1\times 10^{-3}}

C = 3.54 x 10^-12 F

Let q be the charge on each of the electrode

q = C x V

q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C

(b)

As, the battery is disconnected the charge on the electrodes remains same.

(c)

As the battery is connected the voltage is same.

capacitance is change.

As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C

6 0
3 years ago
The 500 series bullet train travels between Tokyo and Hakata, Japan. If it is heading north with a momentum of 13,194,098.64 kg
Hitman42 [59]

Answer:

The mass of the 500 series bullet train is 181.4 tonnes.

Explanation:

The momentum is given by:

p = mv

Where:

m: is the mass =?

v: is the velocity = 261.8 km/h

p: is the momentum = 13194098.64 kg*m/s

By solving the above equation for "m" we have:

m = \frac{p}{v} = \frac{13,194,098.64 kg*m/s}{261.8 km/h*\frac{1000 m}{1 km}*\frac{1 h}{3600 s}} = 181.4 tonnes

Therefore, the mass of the 500 series bullet train is 181.4 tonnes.

I hope it helps you!                

5 0
3 years ago
Suppose an electron and a proton move at the same speed. which particle has a longer de broglie wavelength? suppose an electron
BARSIC [14]
When both particles, the electron and the proton move at the same speed, they may have differences with their de Broglie wavelength, the particle that would have a longer wavelength would be the proton since the wavelength is in direct proportionality with the mass of the particle.
6 0
3 years ago
PLEASE HELP ME A lens with a surface that curves outward like the exterior of a sphere is __________. (Points : 1) reflected ref
yanalaym [24]
Convex.

Concave curved inward (like how a cave foes in) and convex curves outward. Reflected and refracted do not apply to a lens.
3 0
3 years ago
Read 2 more answers
You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Base
shepuryov [24]

Answer:

Explanation:

 For entry of light into tube of unknown refractive index

sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.

sin ( 90 - 25 ) / sin( 90 - C)  = μ

sin65 / cos C = μ

sinC = 1.33 / μ  , where 1.33 is the refractive index of body liquid.

From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

TanC = 1.4675

C= 56°

sinC = 1.33 / μ

μ = 1.33 / sinC

= 1.33 / sin56

= 1.33 / .829

μ = 1.6   Ans

3 0
3 years ago
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