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Gnesinka [82]
3 years ago
10

Which of these is an example of a mechanical wave?

Physics
2 answers:
Oduvanchick [21]3 years ago
8 0

i think the answer is B police siren.

Vikki [24]3 years ago
6 0
The answer to this B police siren
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The acceleration due to gravity on the surface of Mars is gmars = 3.7 m/s2. How much would an 11 kg bag of potatoes weigh on Mar
allsm [11]
Fnet = (mass) (acceleration)
        = 11 kg x 3.7m/s^2
        = 41 N

6 0
3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
Which clue indicates that a speciation has occurred between two populations?
Diano4ka-milaya [45]
I believe the answer is C
4 0
3 years ago
The machine in the figure is ideal and an effort force of
DENIUS [597]

Answer:

1.5 m

Explanation:

Let the distance from the box to the pivot be c.

Let the distance from the pivot to the effort be y.

From the question given above, the following data were obtained:

Effort force (Fₑ) = 7 N

Force of resistance (Fᵣ) = 14 N

Distance from the box to the pivot (c) = 0.75 m

Distance from the pivot to the effort (y) =?

Clockwise moment = Fₑ × y

Anticlock wise moment = Fᵣ × c

Clockwise moment = Anticlock wise moment

Fₑ × y = Fᵣ × c

7 × y = 14 × 0.75

7 × y = 10.5

Divide both side by 7

y = 10.5 / 7

y = 1.5 m

Therefore, the distance from the pivot to the effort is 1.5 m

5 0
2 years ago
A car speeds up from 12.0 m/s to 16.0 m/s in 8.00s what is the acceleration
Lerok [7]

Answer:

0.5m/s²

Explanation:

acceleration =v-u/t

=(16-12)/8

=4/8

acceleration =0.5m/s²

7 0
3 years ago
Read 2 more answers
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