A 10kg block of material is heated by an electric heater rated at 90w lf the temperature of the block rais by 15⁰c and no heat energy last to the surrounding for how many long is the heater switehed on​

(a) The angular acceleration of the centrifuge as it spins up is approximately 9.27 rad/s².

(b) The centrifuge makes approximately 157.08 revolutions as it goes from rest to its final angular speed.

(c) The tangential acceleration of the centrifuge is approximately 1.20 m/s².

To solve this problem, we can use the following equations of rotational motion:

(a) The equation relating angular acceleration (α), time (t), and final angular speed (ω) is:

ω = α * t

We are given the mass (m) of the centrifuge as 4.53 kg, and the radius (r) of the point as 13.00 cm = 0.13 m. The speed (v) of the point is given as 211 m/s. We can use this information to calculate the angular speed (ω) using the formula:

ω = v / r

Substituting the given values, we have:

ω = 211 m/s / 0.13 m ≈ 1623.08 rad/s

Now, substituting the known values into the equation ω = α * t, we can solve for α:

1623.08 rad/s = α * 175 s

α ≈ 9.27 rad/s²

(b) The number of revolutions (N) can be calculated using the formula:

N = ω * t / (2π)

Substituting the known values, we have:

N = 1623.08 rad/s * 175 s / (2π)

N ≈ 157.08 revolutions

(c) The tangential acceleration (at) can be calculated using the formula:

at = α * r

Substituting the known values, we have:

at = 9.27 rad/s² * 0.13 m

at ≈ 1.20 m/s²

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One of the most effective time management strategies that I have used is creating a to-do list. They are successful because they help me to prioritize tasks, remain organized, increase productivity, and reduce stress.

One of the most effective time management strategies that I have used is creating a to-do list.

Creating a to-do list helps me to prioritize tasks and ensures that I do not forget any important tasks.

When creating a to-do list, I ensure that I put the most important tasks at the top of the list and then work my way down.

The to-do list has helped me to organize my work, manage my time effectively and reduce stress.

Another effective strategy that I have used is breaking large tasks into smaller manageable tasks.

When dealing with complex tasks, I break them down into smaller, more manageable chunks.

This helps me to focus on the individual parts of the task, which are easier to handle and manage.

When I focus on smaller tasks, I find it easier to get started, and I gain momentum as I make progress on each small task.

This method has helped me to increase my productivity and reduce the stress that comes with handling large tasks.

In conclusion, the above time management strategies have helped me to manage my time effectively at home, school, and in the workplace.

They are successful because they help me to prioritize tasks, remain organized, increase productivity, and reduce stress.

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Question 12: The correct answer is Ionic, because Lithium is a metal and Chlorine is a nonmetal. Option C) is correct

Question 13: The correct answer is Metallic, because both Copper and Zinc are metals. Option A) is correct

Question 14: The correct answer is 2 valence electrons. Option B) is correct.

Question 15: The correct answer is 5 valence electrons. Option A) is correct.

Question 16: The correct answer is Carbon (C). Option C) is correct

Question 12: The correct answer is Ionic, because Lithium is a metal and Chlorine is a nonmetal. When a metal and nonmetal bond, they form an ionic bond, which involves the transfer of electrons from the metal to the nonmetal. Question 13: The correct answer is Metallic, because both Copper and Zinc are metals. When two metals bond, they form a metallic bond, which involves the sharing of electrons among a lattice of metal atoms. Question 14: The correct answer is 2 valence electrons, because Calcium is in group 2 of the periodic table, and elements in group 2 have 2 valence electrons. Question 15: The correct answer is 5 valence electrons, because Nitrogen is in group 5 of the periodic table, and elements in group 5 have 5 valence electrons. Question 16: The correct answer is Carbon (C), because Carbon is in group 4 of the periodic table, and elements in group 4 have 4 valence electrons.

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Answer:

2 because I did this on before

Answer:

A = 21 î + 28 ĵ

B = 12 î + 5 ĵ

Explanation:

a.

To write A and B in terms of î and ĵ, we need to use the trigonometric ratios and the vector notation

According to the diagram, we have:

tan a = 4/3 tan ß = 5/12

Using the identity tan θ = opposite/adjacent, we can find the x and y components of A and B.

For A, we have:

x component = 35 cos a y component = 35 sin a

Using tan a = 4/3, we can find cos a and sin a by using Pythagoras’ theorem:

cos a = 3/5 sin a = 4/5

Therefore, the x and y components of A are:

x component = 35 cos a = 35 (3/5) = 21 y component = 35 sin a = 35 (4/5) = 28

Using the vector notation, we can write A as:

A = 21 î + 28 ĵ

Similarly, for B, we have:

x component = 13 cos ß y component = 13 sin ß

Using tan ß = 5/12, we can find cos ß and sin ß by using Pythagoras’ theorem:

cos ß = 12/13 sin ß = 5/13

Therefore, the x and y components of B are:

x component = 13 cos ß = 13 (12/13) = 12 y component = 13 sin ß = 13 (5/13) = 5

Using the vector notation, we can write B as:

B = 12 î + 5 ĵ

b.

To show that A + B makes an angle of 45° to the x-axis, we need to find the resultant vector R and its angle θ with the x-axis.

To find R, we can use the vector addition rule :

R = A + B R = (21 î + 28 ĵ) + (12 î + 5 ĵ) R = (21 + 12) î + (28 + 5) ĵ R = 33 î + 33 ĵ

To find θ, we can use the inverse tangent function :

tan θ = y component / x component tan θ = 33 / 33 tan θ = 1

θ = tan^-1(1) θ = 45°

Therefore, A + B makes an angle of 45° to the x-axis.

The bullet fired at an angle of 80° with the horizontal and an initial velocity of 420 m/s can travel up to a height of 825.4 meters and a horizontal distance of 80.1 meters after 2 seconds.

To determine the height and horizontal distance traveled by a bullet fired at an angle of 80° with the horizontal and an initial velocity of 420 m/s after 2 seconds, we can use the equations of motion.

Firstly, we can break down the initial velocity of the bullet into its horizontal and vertical components. The horizontal component remains constant throughout the motion and is given by:

Vx = Vcosθ

where V is the initial velocity and θ is the angle of projection. Substituting the given values, we get:

Vx = 420cos80° = 40.05 m/s (approx.)

The vertical component of the initial velocity can be calculated as:

Vy = Vsinθ

Substituting the given values, we get:

Vy = 420sin80° = 416.95 m/s (approx.)

Now, we can use the following equations of motion to determine the height and horizontal distance traveled by the bullet after 2 seconds:

Vertical motion:

y = yo + Voyt + (1/2)gt^2

where y is the vertical displacement, yo is the initial height (assumed to be zero), Voy is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Substituting the given values, we get:

y = 0 + 416.95(2) - (1/2)(9.8)(2)^2

y = 825.4 m (approx.) Therefore, the bullet can travel up to a height of 825.4 meters after 2 seconds. Horizontal motion:

x = xo + Voxt

where x is the horizontal displacement, xo is the initial horizontal position (assumed to be zero), Vox is the initial horizontal velocity, and t is the time.

Substituting the given values, we get:

x = 0 + 40.05(2)

x = 80.1 m (approx.)

Therefore, the bullet can travel a horizontal distance of 80.1 meters after 2 seconds.

In summary, the bullet fired at an angle of 80° with the horizontal and an initial velocity of 420 m/s can travel up to a height of 825.4 meters and a horizontal distance of 80.1 meters after 2 seconds.

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The power used, given that a force of 20 N is used to push the shopping cart 3.5 m in 2 seconds is 35 W

First, we shall determine the work done in pushing the cart. Details below:

Wd = Fd

Wd = 20 × 3.5

Wd = 70 J

Finally, we shall determine the power used in pushing the cart. Details below:

P = Wd / t

P = 70 / 2

P = 35 W

Thus, we can conclude that the power used is 35 W

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