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vazorg [7]
3 years ago
12

A child attaches a rubber ball to a string and whirls it around in a circle overhead. If the string is 0.2m long and the ball's

speed is 16 m/s, what is the ball's centripetal acceleration?
Physics
1 answer:
AlladinOne [14]3 years ago
4 0

Answer:

The centripetal acceleration, a_c = 1280 m/s^2

Explanation:

Centripetal acceleration is given by the formula:

a_c = \frac{v^2}{r}

Where the speed of the ball, v = 16 m/s

Length of the string, r = 0.2 m

Substituting these parameters into the centripetal equation above:

a_c = \frac{16^2}{0.2}\\a_c = \frac{256}{0.2}\\a_c = 1280 m/s^2

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A charged particle is surrounded by an electric field and a magnetic field
zlopas [31]

Answer:

yeah

Explanation:

electric fields help the charged particles interact

and isn't magnetic field the same as electric field.

7 0
3 years ago
Read 2 more answers
Calculate the magnitude and the direction of the resultant forces​
snow_tiger [21]

answer:

resultant = 127.65 in the positive direction

explanation:

F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N

Fy = 50 sin 50 = 50 × -0.26 = -13

Fx = 40 cos 0 = 40×1 = 40

fx = 55 cos 25 = 55×0.99 = 54.45

Fy = 60 sin 70 = 60 × 0.77 = 46.2

resultant = -13+40+54.45+46.2 = 127.65 in the positive direction

6 0
2 years ago
A stone was dropped off a cliff and hit the ground with a speed of 96 ft/s. What is the height of the cliff? (Use 32 ft/s2 for t
nadezda [96]
The initial velocity of the stone is 0 ft/s. Given the initial velocity (Vi), final velocity (Vf), and acceleration due to gravity (g), the distance may be calculated through the equation,
                                     d = ((Vf)² - (Vi)²) / 2g
Substituting the known values,
                                     d = ((96 ft/s)² - 0))/ (2x32.2)
The value of d is 143.10 ft. 
4 0
3 years ago
Can anyone help me? (physics)
Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0
3 years ago
A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere
dmitriy555 [2]

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

5 0
3 years ago
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