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shutvik [7]
4 years ago
11

When you see Marjorie, she is swinging a bucket of water around in a circle. The speed of the bucket of water does not change as

it moves in the circle. Which of the following statements best explains how you could tell that a net force is constantly acting on the bucket of water?
A. The bucket of water is not at rest.
B. The bucket of water is speeding up or slowing down.
C. The bucket of water began at rest, and you saw it begin to move.
D. The bucket of water's direction is changing, instead of moving in a straight line.
Physics
1 answer:
ser-zykov [4K]4 years ago
4 0

Answer:A

Explanation:

Since the bucket of water  is constantly in motion it implies a force must be acting on it which allows it to change its Position constantly.

A constant gravity Force must be acting on water . In addition to gravity force a centripetal force must be acting on the water inside the bucket.

Bucket must move with certain velocity in order to avoid spilling of water.                    

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The piano ended up 5 meters above the ground.

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Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7
inysia [295]

Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

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The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

4 0
4 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:

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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
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