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3 years ago

Which statement is true about the ideal mechanical advantage of a third-class lever?

Physics

1 answer:

zhannawk [14.2K]3 years ago

6 0

Answer:

Explanation:

The mechanical advantage is always less than 1 because the force needed to move an object is always greater than the weight of the object.

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How is cold front formation different from stationary front formation?

padilas [110]

Answer:

Explanation:

4 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago

Why is the picture above not an accurate representation of charges and their field lines?

pochemuha

Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.

7 0

2 years ago

Read 2 more answers

What is the magnitude of the electric force on an electron in a uniform electric field of strength 2050 n/c that points due east

erastova [34]

Given E=2050 N/C

charge of an electron = 1.6×10^-19 C

USING E = F/Q , electric field = force ÷ charge

2050 = F ÷ (1.6×10^-19)....

make F subject of formula and get the answer

3 0

3 years ago

What is the correct answer?

Anastaziya [24]

Answer:

i think 2 but not sure

Explanation:

4 0

3 years ago

Read 2 more answers