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attashe74 [19]
3 years ago
10

A block lies on a plane raised an angle θ from the horizontal. Three forces act upon the block: F⃗ w, the force of gravity; F⃗ n

, the normal force; and F⃗ f, the force of friction. The coefficient of friction is large enough to prevent the block from sliding
Physics
1 answer:
Verdich [7]3 years ago
5 0

Answer:

YFy = 0 = Ffsinθ + Fncosθ - Fw

Explanation:

From the base of the vector Fn, draw a vertical line. the small angle between this line and Fn is also theta. The component of Fn in the vertical direction is Fncos(theta).

Take a moment to picture extreme cases. Sine is 0 at 0 and 1 at 90. Cosine is 1 at 0 and 0 at 90.

Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.

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REY [17]

Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.

<h3>What is the one-solar mass star?</h3>

A star having a mass equal to the mass of the Sun is called a one-solar mass star.

Its life track shows the luminous intensity as well as the surface temperature.

Learn more about one-solar mass star.

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7 0
2 years ago
Pen and ink manufacturers are asked to submit their new ink formulations to what database?
Ilia_Sergeevich [38]
The international ink library.
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3 years ago
During spring tide, the sun and moon are aligned. Which statement below best describes the tidal range during a spring tide?
e-lub [12.9K]

Answer:

Explanation:

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4 0
3 years ago
Read 2 more answers
A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.
ELEN [110]
Momentum = mass x velocity

Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s

After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2

Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = <span>6.67 m/s

Total kinetic energy before
</span><span>= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J
</span>
<span>Total kinetic energy after
</span>= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J

Total kinetic energy lost during collision
=<span>550 J - 91.73 J
= 458.27 J</span>

8 0
3 years ago
Read 2 more answers
If it actually hits the ground with a speed of 8.50 m/s , what is the magnitude of the average force of air resistance exerted o
kkurt [141]
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8 0
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