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2 years ago

What's a solution that feels slippery

1 answer:

4 0

What are you exactly asking for? A solution that feels slippery? That could mean anything. Please explain a little bit more.

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The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

$\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}$

$\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}$

$\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}$

$\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}$

$\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}$

Work done is defined as

$W = \overrightarrow{F}.\overrightarrow{S}$

$W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k} \right )$

W = -1120 + 450 + 70

W = - 600 J

3 0

2 years ago

ASAP PLS HELP<br> What is chemical potential energy?

mash [69]

Answer:

chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, e.g. in a chemical reaction or phase transition.

Explanation:

5 0

2 years ago

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

2 years ago

What is the mass of the other block?

xeze [42]

Let us consider the tension produced on both the sides of the rope is T.

We have been given that the rope is mass less and the rope is passing through a pulley which is stationary.

$Let\ m_{1} and\ m_{2}\ are\ the\ masses\ of\ two\ blocks$

$Let\ m_{1}\ is\ moving\ in\ vertical\ upward\ direction\ while\ m_{2}\ is\ in\ downward\ \ direction$

$Hence\ m_{2} =4.5\ kg$

The body is moving downward with an acceleration of $\frac{3}{4} g$

As the rope is the same one which is passing over a mass less pulley and connected by two masses,hence, acceleration of each block will be the same in magnitude.

$For\ body\ m_{1}$

Here the tension is acting in vertical upward direction and the weight is acting in vertical downward direction. Here,the body is moving in vertical upward direction. Hence, the net force acting on it is-

$T-m_{1} g=m_{1}a$ [1]

$For\ body\ m_{2}$

Here the tension is acting in vertical upward direction while weight is in vertical downward direction. The body is moving in downward direction. Hence the net force acting on it will be-

$m_{2} g-T=m_{2} a$ [2]