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3 years ago

How can force-time and force-displacement graphs be used to find the impulse or work done?

Physics

1 answer:

balandron [24]3 years ago

7 0

Answer:

A. Area under force-time graph & Area under force-displacement graph

Explanation:

To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.

Impulse = Force x time

Work done = Force x displacement

When we plot a graph of force and time, the area under it is the impulse.

When a graph of force and displacement is plotted, the area under is the work done.

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A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find th

Gre4nikov [31]

Answer:

Assume that $\rm g= 9.81\; N\cdot kg^{-1}$ ; $\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}$ .

Density of the disk: approximately $\rm 2.19\times 10^{3}\; kg\cdot m^{-3}$ .

Weight of the disk: approximately $\rm 245\;N$ .

Buoyant force on the disk if it is submerged under water: approximately $\rm 112\; N$ .

The disk will sink when placed in water.

Explanation:

Convert the dimensions of this disk to SI units:

Diameter: $d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m$ .
Thickness $h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m$ .

The radius of a circle is 1/2 its diameter:

$\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m$ .

Volume of this disk:

$V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}$ .

Density of this disk:

$\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}$ .

$\rho(\text{disk}) >\rho(\text{water})$ indicates that the disk will sink when placed in water.

Weight of the object:

$W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N$ .

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately $\rm 0.0114009\; m^{3}$ of water. The buoyant force on the disk will be:

$\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}$ .

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

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3 years ago

A crowbar having the length of 1.75m is used to balance a load between the fulcrum and the load is 0.5m calculate MA

aliina [53]

Answer:

1.70

Explanation:

substract 1.75-0.5

3 0

3 years ago

Mug<br><br> xx nbbbbnbebe <br> lzu bbt here ju

konstantin123 [22]

Answer:

and a dumb person would say "what's the question" or "I don't get it"

8 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

gayaneshka [121]

Work = force x distance

So we are looking for something related to displacement.

The work done must also be done in the same direction, parallel to the displacement, and therefore in the same direction of the motion as well.

So:

In order to do work, the force vector must be in the same direction as the displacement vector and the motion.

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3 years ago

A 36 kg child sits in a swing supported by two chains, each 2.9 m long. If the tension in each chain at the lowest point is 265

natka813 [3]

Answer:

v = 3.78 m/s

Explanation:

given,

mass of the child = 36 Kg

length of the chain, r = 2.9 m

tension in each chain, T = 265 N

now,

Net force acting in the system

2 T - m g = m a

T is the tension in the chain

a is the acceleration in the normal direction

here, $a = \dfrac{v^2}{r}$

now,

$2 T - m g = m\dfrac{v^2}{r}$

$2\times 265 - 36\times 9.8 = 36\times \dfrac{v^2}{2.9}$

v² = 14.274

v = 3.78 m/s

speed of the child at the lowest point is equal to 3.78 m/s

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3 years ago