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3 years ago

15

Which of the following statements is TRUE?

1 answer:

Marrrta [24]3 years ago

5 0

Answer:

I think C......................

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A ball is dropped off a building and falls past a window that is 2.2m long. If it takes .28s for the ball to cross the window wh

Sidana [21]

0.616 is the distance from the top
of the building to the top of the window.

8 0

2 years ago

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

4 0

3 years ago

005 10.0 points

velikii [3]

Answer:

16.6 N

Explanation:

m = 0.52 kg, v₀ = 0, v = 8.6 m/s, t = 0.27 s

a = (v - v₀)/t

F = ma = m(v - v₀)/t = 0.52 (8.6 - 0)/0.27 = 16.6 N

8 0

2 years ago

Circular Motion A 650-kg car moving at 8.5 m/s takes a turn around a circle with a radius of 48.0 m. Determine the acceleration

ollegr [7]

Answer:

Explanation:

The mass of the car doesn't matter because On a flat curve the mass of the car does not affect the speed at which it can stay on the curve. You would need the mass if you were solving the the centripetal force acting on the car, but not the acceleration.

$a=\frac{v^2}{r}$ and filling in

$a=\frac{(8.5)^2}{48.0}$ and we need 2 significant digits in our answer. That means that

a = 1.5 m/sec²

8 0

3 years ago

A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer

Mrac [35]

Explanation:

The given data is as follows.

$m_{1}$ = 57 kg, $m_{2}$ = 79 kg

$l_{1}$ = 6.5 m, $l_{2}$ = (6.5 - 1.9) m = 4.6 m

(a) The sum of torque ends about far end is as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1}$ = 0

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5$ = 0

T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b) Now, we will calculate the sum about close ends as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}$

$57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5$ = 0

T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.

8 0

3 years ago