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3 years ago

12

in the system to the right, the pulleys are frictionless and the system hangs in equilibrium. Determine the values of each of th

e unknown weights.

1 answer:

irga5000 [103]3 years ago

4 0

Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .

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How do things rust? Is this chemical or mechanical weathering?

IrinaK [193]

I'll say that's mechanical weathering

6 0

3 years ago

Read 2 more answers

True of false efficiency compared the output work to the output force

Lerok [7]

The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>

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3 years ago

A cardinal (Richmondena cardinalis) of mass 3.70×10−2 kg and a baseball of mass 0.144 kg have the same kinetic energy. What is t

Radda [10]

Answer:

$\frac{p_{c}}{p_{b}}\approx 0.507$

Explanation:

Since the cardinal and ball have the same kinetic energy, it is possible to determine the ratio between speeds. (c for cardinal, b for baseball)

$K_{c} = K_{b}$

$\frac{1}{2}\cdot m_{c}\cdot v_{c}^{2}= \frac{1}{2}\cdot m_{b}\cdot v_{b}^{2}$

$\frac{v_{c}}{v_{b}}=\sqrt{\frac{m_{b}}{m_{c}} }$

The ratio is obtained by multiplying each side by $\frac{m_{c}}{m_{b}}$ :

$\frac{p_{c}}{p_{b}}=\frac{m_{c}}{m_{b}}\cdot \sqrt{\frac{m_{b}}{m_{c}} }$

$\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }$

The value of this ratio is:

$\frac{p_{c}}{p_{b}}\approx 0.507$

3 0

3 years ago

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Car accelerates 86km/hr in 4.6s . What is the average acceleration in m/s^2

never [62]

speed = 86km/hr

= 86000/3600

= 23.889m/s

time = 4.6s

acceleration = 23.889/4.6

a = 5.19m/s^2

4 0

3 years ago

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

3 0

1 year ago