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GREYUIT [131]
2 years ago
5

A girl at an airport rolls a ball north on a moving walking that moves east. If the ball’s speed with respect to the walkway is

0.15m/s and the walkway moves at a speed of 1.50m/s, what is the velocity of the ball relative to the ground?
Physics
1 answer:
sweet-ann [11.9K]2 years ago
4 0

We have to add two vectors.

Vector #1: 0.15 m/s north

Vector #2: 1.50 m/s east

Their sum:

Magnitude: √(0.15² + 1.50²)

Magnitude = √(0.0225+2.25)

Magnitude = √2.2725

Magnitude = <em>1.5075 m/s</em>

Direction = arctan(0.15/1.50) north of east

Direction = <em>5.71° north of east</em>

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Buffalo, New York, experienced a snowstorm November 13–21, 2014. Residents refer to the event as “Snowvember.” What was the like
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(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
ddd [48]

Answer:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force exerted by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
2 years ago
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A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the
Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0
2 years ago
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