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3 years ago

Please help me!!!!!!!!!!Which is true about the actual mechanical advantage of a machine?

Physics

2 answers:

Maurinko [17]3 years ago

7 0

Answer:

It is less than the ideal mechanical advantage

Explanation:

Hi, the actual mechanical advange is less than the ideal. This happens because when making calculations and designs of a machine, the scientist or engineer makes a lot of suppositions that doesn't always reflect the reality perfectly.

That difference between the suppositions and the reality is shown in the difference between the actual mechanical advantage and the ideal.

The porcentual difference between this two is what is called the efficiency

Hoochie [10]3 years ago

3 0

B it increases with greater friction

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6 0

3 years ago

Read 2 more answers

The magnitude of the force involved in a certain collision (measured in newtons) is equal to the change in momentum (measured in

Vinvika [58]

Answer:

The time-interval of the collision is $t=\frac{{\Delta}p}{F}$

Explanation:

As given the force is equal to the rate of change of momentum. Mathemticaly this is:

$F=\frac{dp}{dt}.$

We can rearrange this equation to solve for ${\delta}t$ which gives

$t=\frac{{\Delta}p}{F}$

which is our answer.

In words this means the time interval is equal to the momentum change in that interval divided by the force applied that caused this momentum change.

4 0

3 years ago

If it requires 7.0 JJ of work to stretch a particular spring by 1.7 cmcm from its equilibrium length, how much more work will be

Lynna [10]

Answer:

Explanation:

First of all, well calculate the spring constant k

K = 2Ei/x^2

Where Ei = initial work required

x = initial stretch length

k = 2×7/0.017^2 = 48443J/m^2

Now work done in stretching it to 5.3cm (1.7 + 3.6) or 0.053m

EF = kx^2/2

48443 × 0.053^2/2 = 68J

Work done in stretching additional 3.6cm is equal to

68J-7J = 61J

3 0

4 years ago

How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius

mixas84 [53]

The magnitude of static friction is

f = mv ²/r

(i.e. the net force acting on the car parallel to the road points toward the center of the curve)

while the net vertical force must be

∑ F = n - mg = 0

because the car is otherwise in equilibrium. Then

==> n = mg

==> f = µn = µmg = mv ²/r

==> µ = v ²/(rg)

We have

v = 101 km/h ≈ 28.1 m/s

r = 110 m

g = 9.80 m/s²

so that

µ = (28.1 m/s)² / ((110 m) g) ≈ 0.730

6 0

3 years ago

For the electric power transmission system shown in the figure above, what is the ratio ns/np for the step-up transformer? assum

Ann [662]

I am attaching the missing figure

If we assume 100% efficiency, this means that the input power of the transformer must be equal to the output power

We divide output/input

240,000/12,000 = 20

This means that the transformer has a turns ratio of 1:20

ns/np = 20/1 = 20

The voltage is amplified 20 times by the transformer, but this also means that the current decreases by a factor of 20, so the total power remains the same.

7 0

3 years ago