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Mila [183]
2 years ago
7

A wheel is rotating about a fixed axis with constant angular acceleration 3 rad/s². At different moments, its angular speed is -

2 rad/s, 0 , and +2 rad/s. For a point on the rim of the wheel, consider at these moments the magnitude of the tangential component of acceleration and the magnitude of the radial component of acceleration. Rank the following five items from largest to smallest:(a) |at| when ω = -2rad/s, (b) |ar| when ω =-2rad/s, (c) |ar| when ω=0 ,(d) |at| when ω=2 rad/s, and(e) |ar| when ω=2rad/s . If two items are equal, show them as equal in your ranking. If a quantity is equal to zero, show that fact in your ranking.
Physics
1 answer:
meriva2 years ago
8 0

b and e are the largest and equal in magnitude. w*2R = (2)(2)R = 4R

A and d are next. aR = (3rad/s2)R = 3R

c is zero. wR = v = 0; Angular acceleration is zero.

<h3>What is angular acceleration?</h3>
  • The temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second. Therefore, = d d t. Rotational acceleration is another name for angular acceleration.
  • Angular velocity divided by acceleration time can be used to define angular acceleration. (t). As an alternative, use pi times the drive speed (n) divided by the acceleration time (t) times 30. Radians per second squared (Rad/sec2) is the standard SI unit for rotational acceleration resulting from this equation.
  • To calculate angular velocity, we can use one of three formulas. The definition itself provides the first. Theta = position angle, t = time, and w = angular velocity, where w = angular velocity, theta = position angle, and t = time. Angular velocity is the rate of change of an object's position angle with respect to time.
  • The symbol for angular acceleration is, and it is measured in rad/s2, or radians per second square.

If two items are equal, show them as equal in your ranking. If a quantity is equal to zero, show that fact in your ranking:

b and e are the largest and equal in magnitude. w*2R = (2)(2)R = 4R

A and d are next. aR = (3rad/s2)R = 3R

c is zero. wR = v = 0; Angular acceleration is zero.

To learn more about angular acceleration, refer to:

brainly.com/question/20912191

#SPJ4

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The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
Paraphin [41]

(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27  \ m/sec

b)

The maximum height of the projectile is;

\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0
2 years ago
What family of elements has 7 valence electrons?
ivann1987 [24]
Group 17 (VII) (halogens)
4 0
3 years ago
Read 2 more answers
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 110 m above the
zysi [14]

Answer:

The distance is 709.5 m.

Explanation:

Given that,

Speed = 150 m/s

Distance = 110 m

Suppose, How far short of the target should it drop the package?

We need to calculate the time

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

t^2=\dfrac{2s}{g}

Where, g = acceleration due to gravity

t = time

Put the value into the formula

t=\sqrt{\dfrac{2\times110}{9.8}}

t=4.73\ sec

We need to calculate the distance

Using formula of distance

d= vt

Put the value into the formula

d=150\times4.73

d=709.5\ m

Hence, The distance is 709.5 m.

8 0
3 years ago
A 17Kg sitting on a shelf has a potential energy of 350J. How high is the shelf? Round your number to the nearest whole number
saul85 [17]

Answer:

{ \bf{potential \: energy = mass \times gravitation \times height}} \\ 350 = 17 \times 10 \times h \\ 350 = 170h \\ height = 2.06 \: metres

6 0
3 years ago
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