Can you help please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

During what intervals was jenny negatively accelerating

PolarNik [594]

Jenny's acceleration was negative during the intervals of time when her speed was decreasing.

That's really all we can tell, because you really haven't given us any information except her name.

5 0

4 years ago

A man has a weight of 100 Newtons. How much work is done if he climbs 4meters up a ladder?

Aliun [14]

Answer:

400 J

Explanation:

Work is done when a force that is applied to an object moves that object.

The work is calculated by multiplying the force by the amount of movement of an object

W = F * d

here the man has to work against the gravitational field. (against his weight)

F =100 N

Work done = F * d

= 100 * 4

= 400 J

7 0

4 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

so

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

3 years ago

Let's begin by determining the equilibrium position of a seesaw pivot. You and a friend play on a seesaw.Your mass is____

expeople1 [14]

Answer:

1.2 m from the left end

Explanation:

$M$ = mass of the person = 90 kg

$m$ = mass of friend = 60 kg

$r$ = distance of pivot from the person or from the left end

$L$ = length of the seesaw board = 3.0 m

$s$ = distance of pivot from the friend or from right end = L - r = 3 - r

Using equilibrium of torque about the pivot

$M g r = m g s\\(90) r = (60) (3 - r)\\90 r = 180 - 60 r\\90 r + 60 r = 180 \\150 r = 180 \\r = 1.2$

6 0

3 years ago

How will the electrostatic force be changed if the distance between electric charges is reduced one half of the original distanc

GuDViN [60]

I think the answer to your question is A

8 0

3 years ago