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Ilia_Sergeevich [38]
4 years ago
7

for an object that is speeding up in the positive direction, what does the displacement vs. time graph look like?

Physics
2 answers:
Irina18 [472]4 years ago
8 0

Answer:

displacement vs time graph will look like a parabolic shaped graph

Explanation:

As we know that the displacement of an object is given by kinematics equation when object is moving with uniform acceleration

So we know that

d = v t + \frac{1}{2}at^2

now here since velocity is increasing in positive X direction as well as the acceleration is also in same direction then the displacement vs time graph must be a straight line graph

so we will have

slava [35]4 years ago
7 0

Answer:

The speeding up is steady, it is a parabola (a=V*t+(at^2)/2), and give it's an equation in connection to time, at that point, it is conceivable to discover the separation recipe by utilizing more substantial amount mathematics(integrals).

Explanation:

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tigry1 [53]

Explanation:

KE = q

½ mv² = mCΔT

ΔT = v² / (2C)

ΔT = (200 m/s)² / (2 × 236 J/kg/°C)

ΔT = 84.7°C

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3 years ago
List down all the jovian planets in order of increasing distance from the sun
Olenka [21]

Answer:

There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.

Explanation:

The average distance from the sun is listed below in increasing order.

1. Mercury  - It is the most closet planet to Sun, 57 million km

2. Venus  - 108 million km

3. Earth  - 150 million km

4. Mars  - 228 million km

5. Jupiter  - 779 million km

6. Saturn  - 1.43 billion km

7. Uranus  - 2.88 billion km

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8 0
3 years ago
How far will u go if your wheel' diameter 34in?
Lapatulllka [165]
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4 0
3 years ago
Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If
frez [133]

Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2  

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2  

v2 = v1×A1/A2  

and  

r2 = 2×r1

then:

A2 = 4×A1  

so,

v2 = (v1)/4  

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2  

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)  

     = 9.75×10^4 Pa

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Therefore, the pressure in the wider section is 9.8×10^4 Pa

5 0
3 years ago
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