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2 years ago

Its a motion phisic question but im a little confused. Help me fast please.

Physics

1 answer:

Alenkinab [10]2 years ago

3 0

6: Short way: it cannot be 2.5, 3, or 5 because up to 5 seconds it only has positive velocity so it must be moving forwards.

Long Way: Velocity is in m / s, multiply that by time (s) to get m or displacement. From 0->5 you have a triangle under the curve, (1/2)(5)(20) = 50 meters displaced positive, you need to then look when velocity is under the curve and use a similar equation to solve for the area but make the answer negative. Find the point where it equals -50 and that is where it will have returned.

Answer to 6: B

7. I cannot see the problem enough to answer this. Just know if the line is above 0 velocity is positive so it is moving the direction it started, when it goes below 0 velocity is negative so it is moving opposite direction it started.

8. Accelration is change in velocity. Whatever the slope of the velocity graph is acceleration. At t=8 the slope is 0 because it is not going up or down.

Answer to 8: A

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Which of the following is an obstacle to creating computer-based models for tracking a hurricane?

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Answer:

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3 years ago

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3 years ago

During combustion, a fuel’s electromagnetic energy is converted into thermal energy.

swat32

True I believe..................

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2 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

2 years ago

We can see Objects because of

Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

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Hope this helps you out : )

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3 years ago