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3 years ago

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w strong are you? Could you turn a moving car with just your own strength? No way! If a normal driver inputs about 50 N of force

to turn a car, the output force is 3,750 Newtons! What is the MA of the steering wheel?

1 answer:

malfutka [58]3 years ago

3 0

The mechanical advantage of a simple machine is the measure of its amplified force gain.

The mechanical advantage is defined as the force amplified by a machine to the force required to generate such output.

$Mathematically\ mechanical\ advantage\ MA=\frac{F_{o}} {F_{i}}$

$F_{o} \ and\ F_{i}$ are the amplified force and applied force. We may also consider them as output and input force.

In the given question, the force given to the steering wheel is 50 N.

The output force produced by the steering wheel is 3750 N.

Hence the mechanical advantage will be-

$MA=\frac{F_{o}} {F_{i}}$

$=\frac{3750\ N}{50\ N}$

$=75$ [ans]

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Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

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3 years ago

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If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is

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Answer:

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Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C$

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3 years ago

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Drupady [299]

Answer:

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Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

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from (2) we have;

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put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

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3 years ago