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3 years ago

You pull a sled with a package on it across a snow-covered flat lawn. If you

Physics

2 answers:

adelina 88 [10]3 years ago

8 0

Answer:52.5

Explanation:

kumpel [21]3 years ago

4 0

I’m not sure look it up or ask for a tutor on here it’s free

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GuDViN [60]

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When we talk about closed economy, it means that the economy is self-sufficient. At a closed economy, no imports from the outside are brought inside a nation and no exports are sent out as well

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3 years ago

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YES

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HELP!! ME pls which Which of the following graphs correctly shows the relationship between KE and VELOCITY? *

faltersainse [42]

Answer: Option 3.

Explanation:

Formula for kinetic energy is

K.E = (1 / 2) * (m * v ^ 2)

Assuming mass to be constant,

We can see that K.E is proportional to v^2.

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Option 1 and 2 are eliminated because the relationship between kinetic energy and velocity in the graph is shown as linear. We already know from formula that the relationship is not linear.

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3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is: