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2 years ago

How does a second class lever make our work easier

Physics

1 answer:

dmitriy555 [2]2 years ago

7 0

Answer:

In a second class lever, the load is located between the effort and the fulcrum. If the load is closer to the fulcrum than the effort, then less effort will be required to move the load. If the load is closer to the effort than the fulcrum, then more effort will be required to move the load.

Explanation:

from what i learned, if its farther away from the load, its easier to lift, like a wheel barrel

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Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

$F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}$

The electrostatic force between them is

$F_{e}=\frac{Kq^{2}}{d^{2}}$

According to the question

1 % of Fg = Fe

$0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}$

$2.927 \times 10^{35}=9 \times10^{9}q^{2}$

$3.25 \times 10^{25}=q^{2}$

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0

2 years ago

Problems - Show all work.

Fittoniya [83]

Answer:

21s

Explanation:

Given parameters;

Radius = 10m

Speed or velocity = 3m/s

Unknown:

Period = ?

Solution:

To solve this problem, use the expression:

v = $\frac{2\pi r}{T}$

r is the radius

T is the unknown

Input the parameters and solve for T;

3 = $\frac{2 x \pi x 10}{T}$

62.84 = 3T

T = 21s

4 0

2 years ago

The pressure in a traveling sound wave is given by the equation ΔP = (1.78 Pa) sin [ (0.888 m-1)x - (500 s-1)t] Find (a) the pre

frozen [14]

Answer:

a) $P_m=1.78\ Pa$

b) $f=79.5775\ Hz$

c) $\lambda=7.076\ m$

d) $v=563.06\ m.s^{-1}$

Explanation:

<u>Given equation of pressure variation:</u>

$\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]$

We have the standard equation of periodic oscillations:

$\Delta P=P_m\ sin\ (kx-\omega.t)$

<em>By comparing, we deduce:</em>

(a)

amplitude:

$P_m=1.78\ Pa$

(b)

angular frequency:

$\omega=2\pi.f$

$2\pi.f=500$

∴Frequency of oscillations:

$f=\frac{500}{2\pi}$

$f=79.5775\ Hz$

(c)

wavelength is given by:

$\lambda=\frac{2\pi}{k}$

$\lambda=\frac{2\pi}{0.888}$

$\lambda=7.076\ m$

(d)

Speed of the wave is gives by:

$v=\frac{\omega}{k}$

$v=\frac{500}{0.888}$

$v=563.06\ m.s^{-1}$

8 0

2 years ago

Of gold, oxygen, iron, and sulfur, which is a renewable source?

Ulleksa [173]

Oxygen bc plants go through photosynthesis which keeps producing more oxygen

6 0

3 years ago

Read 2 more answers

A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when

lana66690 [7]

Answer:

8.8 kN

Explanation:

V = 2 m³, W = 40 kN, SG = 1.59

Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN

So the weight becomes 40 - 31.2 = 8.8 kN

3 0

2 years ago

How does a second class lever make our work easier​

How does a second class lever make our work easier