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3 years ago

A certain mountain road is inclined 3.1° with respect to the horizon. what is the change in altitude of the car as a result of i

ts traveling 2.90 km along the road?

Physics

1 answer:

BigorU [14]3 years ago

7 0

$Given:\\ \alpha =3.1^\circ\\l=2.90km\\\\Find:\\h=?\\\\Solution:\\\\h=l\sin\alpha \\\\h=2.90km\cdot \sin 3.1^\circ=2.90km\cdot 0.054=0,1566km=156.6m$

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

elena-14-01-66 [18.8K]

The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

c is speed of light
E₀ is the peak electric field
B₀ is the peak magnetic field

B₀ = E₀ / c

B₀ = (2.9) / (3 x 10⁹)

B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

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2 years ago

The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?

Bas_tet [7]

Answer:

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Explanation:

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2 years ago

A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of

jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

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3 years ago

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Your best bet is to go with D.

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2 years ago

Read 2 more answers

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis

Elza [17]

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1 ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

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3 years ago