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1 year ago

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A student did 24 J of work on a chair. She applied a force of 12 N and moved the chair 2 m. What else do you need to know to det

ermine the amount of power used?

1 answer:

ANEK [815]1 year ago

4 0

Answer:

time

Explanation:

power = work done ÷ time

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A playground is on the flat roof of a city school, 6.2m above the street below. The vertical wall of the building is h=7.30m hig

olya-2409 [2.1K]

Answer:

a. 18.13m/s

b. 0.84m

c. 2.4m

Explanation:

a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that

t=distance /speed

$speed v_{x}=vcos\alpha=24/2.2=10.9m/s\\V=10.9/cos53^{0}\\V=18.13m/s$

Hence the speed at which ball was lunched is 18.13m/s

b. from the equation

$y=v_{y}t-\frac{1}{2}gt^{2}\\ v_{y}=18.13sin53=14.48m/s\\\\y=(14.48*2.2)-\frac{1}{2}9.8*2.2^{2}\\y=8.14m\\$

the vertical distance at which the ball clears the wall is

y=8.14-7.3=0.84m

c. the time it takes the ball to reach the 6.2m vertically

$6.2=14.47t-4.8t^{2}\\\\14.47t-4.8t^{2}-6.2=0\\\using\\t=-b±\frac{\sqrt{b^{2}-4ac} }{2a}\\ where \\a=-4.8, b=14.47t, c=6.2\\hence t=2.4secs$

the horizontal distance covered at this speed is

$y+24=(18.13cos53)2.4\\y=26.186-24\\y=2.12m$

4 0

2 years ago

I need help on this question!! Please answer asap!

valentina_108 [34]

Answer:

hub moon merc saturn sun galaxy deep field

Explanation:

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1 year ago

A fence 8 ft high (w) runs parallel to a tall building and is 24 ft (d) from it. Find the length (L) of the shortest ladder t

AveGali [126]

Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

$Hypotenuse^2 = opposite^2 + adjacent^2$

The hypotenuse is the length of the ladder.

$Hypotenuse = \sqrt{opposite^2 + adjacent^2}$

$L^2 = BC^2 + (24 + AE)^2$ ........1

Triangle ABC is similar to triangle AEF, hence:

$\frac{BC}{8} = \frac{AE + 24}{AE}$

BC = $\frac{8(AE + 24)}{AE}$ .................2

Substitute 2 into 1

$L^2$ = $(\frac{8(AE + 24)}{AE})^2$ + $(24 + AE)^2$

Let AE = x

$L^2$ = $(\frac{8x + 192}{x})^2 + (24 + x)^2$

= $(8 + \frac{192}{x})^2 + (24 + x)^2$

Minimize L with respect to x.

2 $\frac{dL}{dx}$ = $2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)$

=

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2 years ago

A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque abo

kipiarov [429]

Answer:

$\tau =37.34\ N m$

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

$\tau = \vec{r} \times \vec{F}$

$\tau = r \times F cos \theta$

$\tau = 0.55 \times 8 \times 9.8 \times cos 30^0$

$\tau =37.34\ N m$

torque about his shoulder join is equal to $\tau =37.34\ N m$

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3 years ago

6. And what's up with ChiN?

n200080 [17]

I don’t know what’s up with chin man ??

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1 year ago