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2 years ago

How does an inclined plane affect the effort needed to move a load vertically?

1 answer:

5 0

If we pull an object vertically upwards then we need to apply a force which is equal in the magnitude of the weight of the object

$F = mg$

now when we pull the same object upwards along an inclined plane with angle then we require a force which will balance the component of weight along the inclined

so it is given as

$F' = mgsin\theta$

so as if we compare the two forces we can say that since the value of sine is always less than 1 for an angle less than 90 degree

so in the 2nd case when we pull the object along the inclined plane it will require less effort

so correct answer is

<em>A. reduce effort</em>

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Genrish500 [490]

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2 years ago

What is a 7 letter word that describes the relationship between frequency and period?

adoni [48]

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2 years ago

A 6.00 g lead bullet traveling at420 m/s is stopped by a large

Westkost [7]

Answer:

The increase in temperature of the bullet is 351.1 kelvin

Explanation:

First, we should find the kinetic energy of the bullet is:

$K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}$

with m the mass and v the velocity.

$K=529.2 J$

Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:

$Q=\frac{K}{2}= 264.6\,J$

To know what the increase in temperature is, we should use specific heat of lead:

$c=125.604 \frac{J}{kg\.K}$

The equation that relates specific heat, change in temperature and mass is:

$Q=cm\varDelta T$

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$\varDelta T= 351.10\, K$

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2 years ago

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$