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2 years ago

14

How does a second class lever make our work easier

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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If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take

kifflom [539]

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

8 0

3 years ago

A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

My name is Ann [436]

Answer:

a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C , c) U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

C = k e₀ A / d

Let's calculate

C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

C = 4,012 10⁻¹⁴ F

b) let's look the charge

C = Q / ΔV

Q = C ΔV

Q = 4,012 10⁻¹⁴ 400

Q = 1.6 10⁻¹¹ C

c) The stored energy

U = ½ C ΔV²

U = ½ 4,012 10⁻¹⁴ 400²

U = 3.21 10⁻¹¹ J

4 0

2 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago

Which of the following is not a constant or control for this experiment?

SCORPION-xisa [38]

D.the amount of time in sun

8 0

2 years ago

Read 2 more answers

What is the potential difference (voltage) if the resistance of 20 ohms produces a current of 10 amperes?

JulijaS [17]

Answer:

200 volts

Explanation:

Ohm's Law: V = IR

I = 10 A

R = 20 ohms

V = (10)(20) = 200 volts

8 0

2 years ago