The tip of a fan blade is 0.61 m from the center of the fan. The fan turns at a constant speed and completes 2 rotations every 1.0 second. What is the centripetal acceleration of the tip of the fan blade? Option 1: 6.0 m/s² Option 2: 48 m/s² Option 3: 53 m/s² Option 4: 96 m/s²

Answers

Answer 1

Answer:

Answer: Here's my answer, I made it step-by-step so you can understand it! <3

Explanation:

To find the centripetal acceleration of the tip of the fan blade, we can use the formula for centripetal acceleration:

a = (v^2) / r

where:

a is the centripetal acceleration,

v is the linear velocity, and

r is the radius of the circular path.

Given that the fan completes 2 rotations every 1.0 second, we can find the angular velocity (ω) using the formula:

ω = (2π * n) / t

where:

ω is the angular velocity,

π is a constant (approximately 3.14),

n is the number of rotations (2),

and t is the time taken (1.0 second).

Substituting the values into the formula, we have:

ω = (2π * 2) / 1.0 = 4π rad/s

Next, we can calculate the linear velocity (v) using the formula:

v = r * ω

Substituting the given radius value (0.61 m) and the angular velocity we found earlier, we have:

v = 0.61 * 4π = 2.44π m/s

Finally, we can calculate the centripetal acceleration (a) using the formula:

a = (v^2) / r

Substituting the linear velocity and the radius, we have:

a = (2.44π)^2 / 0.61 = 5.88π^2 / 0.61 ≈ 96 m/s²

Therefore, the centripetal acceleration of the tip of the fan blade is approximately 96 m/s² (Option 4).

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Which type of bond will sulfer (S) and nitrogen (N) form?a. metallic bond
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Answer:

In ionic compounds, electrons are transferred between atoms of different elements ... For example, two hydrogen atoms bond covalently to form an H2 molecule; each ... is released when one mole of H2 molecules forms from two moles of H atoms: ... atom in each pair is more electronegative: (a) N or P. (b) N or Ge. (c) S or F.

Explanation:

In which layer does erosion take place? And why?

Answers

In the Earths Crust. Because, the crust includes our land and, erosion occurs on land.

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Hope this concluded your answer and was useful!

____________ occurs when an energy source transfers heat directly to another object through space; an example would be an object becoming warm by sitting in the sunshine. a. conduction
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Answers

the answer is Radiation

Answer:

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Explanation:

An 81 kg stuntman jumps from the top of a building 29 m above a catching net. Assuming that air resistance exerts a 100 N force on the stuntman as he falls, determine his velocity just before he hits the net

Answers

In order to solve this, we can get all tangled up in acceleration,
or we can just add up the energy budget.

-- Gravitational potential energy = (mass) (grav accel) (height above something)

-- 29 m above the net, his potential energy is (81 x 9.8 x 29) = 23,020 joules

-- All the way down, air resistance exerts 100 N of force against him.
The energy burned up by air resistance is the work done = 100 x 29 = 2,900 joules.

-- The energy he has left when he hits the net is (23,020 - 2,900) = 20,120 joules.

-- When he hits the net, all of his energy is kinetic energy . . . (1/2) (m) (v²)

(1/2) (m) (v²) = 20,120

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v² = 20,120 / 40.5

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A block of mass m 5.00 kg ispulled along a horizontal frictionless floor by a cord
that exerts a force of magnitude F 12.0 N at an angle
u 25.0°. (a) What is the magnitude of the
block’s acceleration? (b) The force magnitude F is
slowly increased. What is its value just before the
block is lifted (completely) off the floor? (c) What is
the magnitude of the block’s acceleration just before
it is lifted (completely) off the floor?

Answers

a.
the force in the direction of movement
Fx = F x cosA*
Fx = 12 x cos25*
Fx = 10.88
m x a = 10.88
a = 10.88/5
a= 2.18 m/s^2
b.
when the block will be lifted off the floor and when the vertical component
Fy = mg
Fy = F x sin25*
Fy= 5 x 9.8
Fy= 115.94 N
c.
if Fx = ma
a = 115.94/5
a= 23.19 m/s^2
hope it helps

Final answer:

The magnitude of the block's acceleration is roughly 2.18 m/s². The force required just before the block is lifted off the floor is equal to the weight of the block, 49.05 N. Just before it is lifted, the block's acceleration is still due to the horizontal component of the force.

Explanation:

This problem involves physics concepts related to forces and acceleration of an object on a frictionless surface. The key to solving it involves understanding the relationship between mass, force, and acceleration (Newton's second law) and the concept of vertical and horizontal components of a force.

(a) The magnitude of the block's acceleration can be calculated using the horizontal component of the force (Fh = F cos θ) and Newton's second law (F = ma). So, a = Fh / m = (F cos θ) / m = (12.0 N cos 25) / 5.00 kg ≈ 2.18 m/s².

(b) The force F required to lift the block is equal to the weight of the block mg. Thus, F = m*g = 5.00 kg * 9.81 m/s² = 49.05 N.

(c) Just before the block is lifted off the floor, it is still in contact with the floor so the normal force is not zero. As such, there is no vertical acceleration, so the magnitude of the block's acceleration is still solely due to the horizontal component of F.

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In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running over the edge of the cliff above Ohio's Cuyahoga River in (Figure 1) , which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft (≈ 6.7 m) across while falling 20 ft (≈ 6.1 m).What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

Express your answer to two significant figures and include the appropriate units.

Answers

The minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Further explanation:

As Captain Sam Brady jumps from the cliff, he moves in two dimension under the action of gravity.

Given:

The height of free fall of the captain Brady is $20\text{ ft}$ or $6.1\text{ m}$ .

The horizontal distance moved by the captain Brady is $22\text{ ft}$ or $6.7\text{ m}$ .

Concept:

The time required to free fall of a body can be calculated by using the expression given below.

$\left( { - s}\right)=ut-(1)/(2)g{t^2}$ ……. (1)

The displacement is considered negative because the captain is moving in vertically downward direction.

Here, $s$ is the distance covered by the body in free fall, $u$ is the initial velocity of the object, $g$ is the acceleration due to gravity and $t$ is the time taken in free fall of a body.

As the Caption jumps off the cliff, he has his velocity in the horizontal direction. The velocity of the captain in vertical direction is zero.

Substitute $0$ for $u$ in the equation (1) .

$s=(1)/(2)g{t^2}$

Rearrange the above expression for $t$ .

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$ …… (2)

Converting acceleration due to gravity in $\text{ft}/\text{s}^2$ .

$\begin{aligned}g&=\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {\frac{{1.0\,{\text{ft/}}{{\text{s}}^{\text{2}}}}}{{0.305\,{\text{m/}}{{\text{s}}^{\text{2}}}}}} \right) \n&=32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}} \n \end{aligned}$

Substitute $20\text{ ft}$ for $s$ and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (2) .

$\begin{aligned}t&=\sqrt {\frac{{2\left( {20\,{\text{ft}}} \right)}}{{\left( {32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}} \right)}}} \n&=1.116\,{\text{s}} \n \end{aligned}$

Therefore, the time taken by captain to free fall a height $20\text{ ft}$ is $1.116\text{ s}$ .

In the same time interval captain has to move $22\text{ ft}$ in horizontal direction. The acceleration is zero in horizontal direction. So, the velocity will be constant throughout the motion in the horizontal direction.

The distance travelled by captain in the horizontal direction is given by,

$x=v\cdot t$

Rearrange the above expression for $v$ .

$\boxed{v=(x)/(t)}$ …… (3)

Here, $x$ is the distance travelled in horizontal direction, $v$ is the velocity of the captain and $t$ is the time.

Substitute $22\text{ ft}$ for $x$ and $1.116\text{ s}$ for $t$ in equation (3) .

$\begin{aligned}v&=\frac{{22\,{\text{ft}}}}{{1.116\,{\text{s}}}} \n&=19.71\,{\text{ft/s}} \n \end{aligned}$

Thus, the minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

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Keywords:

Free fall, projectile, gravity, 1780, Brady’s, leap, Captain, Sam Brady, US, continental army, enemies, Ohio’s, Cuyahoga river, 22 ft, 6.7 m, 20 ft, 6.1 m, minimum speed, run off, edge, cliff, safely, far side, river, 19.71 ft/s, 6 m/s, 6 meter/s, 5.99 m/s, 599.8 cm/s.

Final answer:

Using the principles of projectile motion from Physics, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to reach the far side of the river.

Explanation:

This problem can be solved using basic Physics, specifically projectile motion. Here, Captain Sam Brady had to run off the edge of the cliff to make it safely to the far side of the river which is 22 ft away while falling 20 ft down. We assume that he jumps horizontally (i.e., his initial vertical velocity is 0).

Firstly, we calculate the time for the vertical fall. Using the equation t = sqrt (2h/g) where h is height and g is the acceleration due to gravity (32.2 ft/s²), we get time t ≈ 1.12s (rounded to two significant figures).

Next, we can use this time to figure out his initial horizontal velocity needed. The equation v = d/t where v is velocity, d is distance, and t is time gives us v ≈ 19.64 ft/s (rounded to two significant figures).

So, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to make it safely across the river.

Learn more about Projectile Motion here:

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