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ANEK [815]
3 years ago
5

In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff

ort if it losses are negligible​
Physics
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

Force ratio = 2.5

Distance of load from the fulcrum = 0.5 m

Distance of effort =.?

The distance of the effort from the fulcrum can be obtained as illustrated below:

Force ratio = Distance of effort / Distance of load

2.5 = Distance of effort / 0.5

Cross multiply

Distance of effort = 2.5 × 0.5

Distance of effort = 1.25 m

Therefore, the distance of the effort from the fulcrum is 1.25 m

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3 years ago
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of
ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

v^2=u^2+2gh

h =\dfrac{v^2}{2g}....(I)

For second half,

We need to calculate the time

Using equation of motion

h =ut+\dfrac{1}{2}gt_{2}^2

h=0+\dfrac{1}{2}gt_{2}^2

t_{2}=\sqrt{\dfrac{2h}{g}}

Put the value of h from equation (I)

t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}

t_{2}=\dfrac{v}{g}

According to question,

t_{1}+t_{2}=10

t_{1}=t_{2}

Put the value of t₁ and t₂

\dfrac{v}{g}+\dfrac{v}{g}=10

\dfrac{2v}{g}=10

v=\dfrac{10\times g}{2}

Here, g = 10

The initial velocity is

v=\dfrac{10\times10}{2}

v=50\ m/s

Hence, The initial velocity is 50 m/s.

3 0
3 years ago
A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

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p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

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3 years ago
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The SI unit for acceleration is m/s2 ( D)

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