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lys-0071 [83]
3 years ago
11

A toy rotates at a constant 5rev/min. is its angular acceleration positive, negative, or zero?

Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

Its angular acceleration is zero.

Explanation:

If the angular velocity of an object (in this case the toy) is constant, then its angular acceleration will be zero. Why? Because angular acceleration is the time rate change of angular velocity. Since there is no change, this brings the answer to zero.

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The brand of car wax
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A motorcycle is stopped at a stop light. When the light turns green it
ss7ja [257]

Answer: 18.9 m

Explanation:

i did the kinematic equation & found the answer.

8 0
3 years ago
A jet can travel the 6000 km distance between washington,
Andrei [34K]
<span> d = r*t is the basic distance equation
 d = 6000 km
 t with the tail wind = 6 hr
 r with the tail wind = speed of the plane + wind speed = s + w
 t with the head wind = 7.5 hr
 r with the head wind = speed of the plane - wind speed = s-w
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 w = 100
 Check the anwer by calculating the return trip.
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 Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>
6 0
3 years ago
In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

a_r=\omega^2R

Plug in the given values and solve for a_r. This gives,

a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2

Now, tangential acceleration is given by the formula:

a_t=R\alpha

Plug in the given values and solve for a_t. This gives,

a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

a_{Total}=\sqrt{(a_r)^2+(a_t)^2}

Plug in the given values and solve for total acceleration, a_{Total}. This gives,

a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0
3 years ago
A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.0062
galina1969 [7]

Answer:

Force on the proton will be .73\times 10^{-14}N

Explanation:

We have given speed of proton is 20.3% of speed of light

Speed of light c=3\times 10^8m/sec

So speed of proton v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec

Magnetic field B = 0.00629 T

Charge on proton q=1.6\times 10^{-16}C

Angle between velocity and magnetic field \Theta =137^{\circ}

Force on the proton is equal to F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N

4 0
3 years ago
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