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2 years ago

9

How does 3rd class lever make our work easier

2 answers:

Bogdan [553]2 years ago

8 0

Answer:

the third class lever makes our work easier as effort is located at the middle, load at first and fore is applied at the buttom. which helps in accelerating work.

hope it is helpful to you

sineoko [7]2 years ago

7 0

In a third class lever, the effort is located between the load and the fulcrum. If the fulcrum is closer to the load, then less effort is needed to move the load. If the fulcrum is closer to the effort, then the load will move a greater distance. ... These levers are useful for making precise movements.

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2 years ago

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

Bond [772]

Answer:

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

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A car of mass 800kg travels a distance of 40m at constant speed in a duration of 2.0s. The car exerts a forward force of 15kN.

Alex17521 [72]

W = F × s

W = 15kN × 40 m

W = 15.000 N × 40 m

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Answer:

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Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

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Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

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