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pochemuha
3 years ago
12

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.50.(a) What is the o

bject’s acceleration between 0 and 4.0 s? (b) What is the object’s acceleration between 4.0 s and 9.0 s? (c) What is the object’s acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x = 0? (f) What is the final position x of the object at t = 18.0 s? (g) Through what total distance has the object moved between t = 0 and t = 18.0 s?
Physics
1 answer:
bezimeni [28]3 years ago
8 0
Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question
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An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00
SSSSS [86.1K]

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

7 0
3 years ago
Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra
RSB [31]

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

                                                  F = ma

   T     -   Wt ||      -         Ff            = m(0)

  mg   -  Μgsinθ -      μΜgcosθ    = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0        

                                                 m = 13(sin35 + 0.36cos35) / 9.8

                                                 m = 1.15205... ≈ 1.2 kg

5 0
2 years ago
PLEASE HELP QUICK I NEED IT TO PASS
BlackZzzverrR [31]

Answer:

Explanation:

As the contour lines have roughly the same spacing but the actual topography is much steeper, the lines on the mountainous map represent a larger vertical spacing than the lines on the gradual hills.

3 0
2 years ago
All sound waves need a(n) _______________________________________ to travel through. 4. _____________________ and ______________
Paha777 [63]

Answer:

Material medium

compressions and rarefactions

Explanation:

A sound wave is an example of a mechanical wave. All mechanical waves require a material medium for propagation. The medium for the propagation of sound is air. This is the reason why, if you cover your mouth, it will be difficult for another person to hear whatever you are saying.

Sound is also a longitudinal wave. Longitudinal waves are described in terms of compressions and rarefactions. Compressions refer to areas where air molecules crowd together while rarefactions refer to areas where the air molecules spread out.

4 0
3 years ago
A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L
GaryK [48]

Answer:

a) L₂ = 0.676 m

b) σ₁ = 2.28*10⁸ N/m²

σ₂ = 9.62*10⁸ N/m²

c) ε₁ = 0.00253678

ε₂ = 0.00457875

Explanation:

Given info

L₁ = 1.22 m

A₁ = 2.19 cm² = 2.19*10⁻⁴ m²

L₂ = ?

A₂ = 0.52 cm² = 0.52*10⁻⁴ m²

P = 5.00*10⁴ N

E₁ = 9*10¹⁰ N/m²

E₂ = 2.1*10¹¹ N/m²

In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that

ΔL₁ = ΔL₂   ⇒  P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)

⇒  L₂ = A₂*E₂*L₁ / (A₁*E₁)

⇒  L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)

⇒  L₂ = 0.676 m

The stress in the brass rod is obtained as follows

σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²

The stress in the niquel rod is obtained as follows

σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²

The strain in the brass rod is obtained as follows

σ₁ = E₁*ε₁    ⇒   ε₁ = σ₁ / E₁

⇒   ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678

The strain in the niquel rod is obtained as follows

σ₂ = E₂*ε₂    ⇒   ε₂ = σ₂ / E₂

⇒   ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875

3 0
3 years ago
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