All categories

3 years ago

| 'The force needed to push a load up an inclined

Physics

1 answer:

dexar [7]3 years ago

3 0

Explanation:

Since less effort is needed in lifting a load to a higher level by moving over an inclined plane as compared to that in lifting the load directly, an inclined plane acts as a force multiplier. This is because the mechanical advantage of an inclined plane is always greater than 1.

You might be interested in

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

3 years ago

Please help its URGENT!!

Greeley [361]

Answer:

1 D ,2 B ,5 C ,3 A ,4 E i dont know why you gotta have 20 words to answer it but yeah

6 0

2 years ago

Read 2 more answers

Forces that act in equal and opposite directions on an object

Akimi4 [234]

These are known as balanced forces because they will not change the motion of the object, and it will remain at rest unless forces become unbalanced- meaning they would be unequal and not opposing.

5 0

3 years ago

An object is thrown upward from the top of a 144-foot building with an initial velocity of 128 feet per second. The height h of

ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

0 = -16t²+128t+144

16t²-128t-144= 0

$t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s$

Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0

3 years ago

Campaign against crime

erica [24]

Answer:

ok what do you need to know about ''campaign against crime''?

Explanation:

8 0

2 years ago