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erik [133]
3 years ago
12

A Frisbee of radius 0.15 m is accelerating at a constant rate from 7.1 revolutions per second to 9.3 revolutions per second in 6

.0 s. What is its angular acceleration?
Physics
1 answer:
Greeley [361]3 years ago
7 0

Answer: 17.03 rad/s^2

Explanation: One of the equation of motion that defines a circular motion with constant acceleration is given below as

ω = ω0 + αt

Where ω = final angular velocity = 9.3 rev/s

ω0 = initial angular velocity = 7.1 rev/s

α = angular acceleration = ?

t = time taken = 6.0s

By substituting the parameters, we have that

9.3 = 7.1 +α(6)

9.3 - 7.1 = 6α

16.3 = 6α

α = 16.3/6

α = 2.71 rev/s^2

We can also give the answer in rad/s^2 ( another unit of angular acceleration) by multipying by 2π

Hence, α = 2.71 × 2π , where π = 3.142

α = 17.03 rad/s^2

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TiliK225 [7]

Answer:

a) By v^2 = u^2 + 2as => a= 70291.70.

(b)By v = u + at => t= 1.58 ms.

(c)By v^2 = u^2 - 2gh => H = 46045.92 m.

Explanation:

a) By v^2 = u^2 + 2as

(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70

(b)By v = u + at

950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms

(c)By v^2 = u^2 - 2gh

0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m

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2 years ago
Goldberg's sleigh currently runs at 203mph, but he needs it to reach 400mph with all the packages he has to deliver.
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Answer:

c

Explanation:

6 0
3 years ago
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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
A density triangle is a tool that can be used to find
Anna [14]
The answer is D
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a horse runs with an initial velocity of 11m/s and slows to 5.2 m/s over a time interval of 3.1 s what is the horse's average ac
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Answer:

a = change in v / change in time

= (5.2 - 11) / 3.1

= -1.87 m/s^2

Explanation:

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3 years ago
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