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3 years ago

What mountains range can be seen from vancouver wa?

Physics

2 answers:

kirill115 [55]3 years ago

8 0

The North Shore mountains

Amanda [17]3 years ago

8 0

<span>Mt. Si Mt. Olympus Mt. St. Helen's Mt. Baker Mt. Adams <span>Mt. Rainier. those are the six mountains and the one you can see is Mt. Rainier </span></span>

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People who hunt animals, go fishing, and gather fruits, nuts, roots, berries, and other parts of plants that can be used as food

Masja [62]

Native American I think in not sure

7 0

3 years ago

Write 45.670,000 with significant figure

weqwewe [10]

4.467*10^7 is the answer

5 0

3 years ago

The tub of a washer goes into its spin-dry cy-cle, starting from rest and reaching an angularspeed of 3.1 rev/s in 10.7 s.At thi

neonofarm [45]

Answer:

35 revolutions

Explanation:

t = Time taken

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Number of rotation

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3-0}{10.7}\\\Rightarrow \alpha=0.28037\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{3.1^2-0^2}{2\times 0.28037}\\\Rightarrow \theta=17.13806\ rev$

Number of revolutions in the 10.7 seconds is 17.13806

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3.1}{11.2}\\\Rightarrow a=-0.27678\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3.1^2}{2\times -0.27678}\\\Rightarrow \theta=17.36035\ rev$

Number of revolutions in the 11.2 seconds is 17.36035

Total total number of revolutions in the 21.9 second interval is 17.13806+17.36035 = 34.49841 = 35 revolutions

3 0

4 years ago

Why doesn't the motor work?

exis [7]

Answer:

Explanation: its weird

8 0

3 years ago

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

Rzqust [24]

Answer:

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

= 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0

3 years ago