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3 years ago

Helppppppppppppppppppp

Physics

1 answer:

hichkok12 [17]3 years ago

4 0

Answer:

100 million step

Explanation:

wish it will help u

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Three particles are placed in the xy plane. A 50-g particle is located at (3, 4) m, and a 40-g particle is positioned at ( 2, 6)

oksano4ka [1.4K]

Answer:

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Explanation:

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3 years ago

..................................<br>.

vivado [14]

Answer: there is 36 dots

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3 years ago

Classify the following as alkali metals, alkaline earth metals, transition elements, or inner transitional elements: calcium, go

Digiron [165]

Alkali metals : sodium , potassium
alkaline earth : magnesium , calcium
the rest are transition elements... i don't know about "inner transition"

6 0

3 years ago

Read 2 more answers

hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward

dimulka [17.4K]

For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

t = 5 m / 6.65 m/s

t = 0.75 s

Result:

The time will be <u>0.75 seconds.</u>

7 0

3 years ago

Read 2 more answers

A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinde

Romashka [77]

Answer:

(a)10.5 rad/s2

(b) 20.9 rev

Explanation:

As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law

T = mg = 53*9.81 = 519.93 N

Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is

To = TR = 519.93 * 0.36 = 187.17 Nm

This solid cylinder would have a moment of inertia around it's rotating axis of:

$I = \frac{mR^2}{2} = \frac{275 * 0.36^2}{2} = 17.82kgm^2$

(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder

$\alpha = \frac{To}{I} = \frac{187.17}{17.82} = 10.5 rad/s^2$

(b) With such constant angular acceleration, the angle it would make after 5s is

$\theta = \frac{\alphat^2}{2} = \frac{10.5*5^2}{2} = 131.3 rad$

Since each revolution equals to $2\pi rad$ of angle, we can calculate the number of revolution it makes

$\frac{\theta}{2\pi} = \frac{131.3}{6.28} \approx 20.9 rev$

(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad

$\theta R = 131.3*0.36 = 47.27 m$

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3 years ago