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Butoxors [25]
3 years ago
6

During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h

as an angular speed of 12.0 rad/s. If the distance between the top of the racket and the shoulder is 1.30 m, find the magnitude of the total acceleration of the top of the racket.
Physics
1 answer:
QveST [7]3 years ago
3 0

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

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How do intermolecular forces differ from intramolecular forces
kicyunya [14]

Answer:

Explanation:

Intramolecular forces is a strong bond that helps to bond atoms together while intermolecular forces are weak bond that are present between molecules.

8 0
3 years ago
A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal
zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0
3 years ago
A boy kicks a football from ground level. The ball takes 3 seconds to reach its maximum height. What is the angle of the initial
ruslelena [56]
<h2>The angle of the initial velocity with respect to the horizontal is 85.14°</h2>

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

     Acceleration, a = -9.81 m/s²

     Final velocity, v = 0 m/s    

     Time, t = 3 s

     Substituting

                      v = u + at  

                      0 = u + -9.81 x 3

                      u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

        Displacement, s = 15 m

        Acceleration, a = 0 m/s²  

        Time, t = 6 s      

     Substituting

                      s = ut + 0.5 at²

                      15 = u x 6 + 0.5 x 0 x 6²

                      u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

              Initial vertical velocity = rsinθ = 29.43 m/s

              Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

              \frac{rsin\theta }{rcos\theta }=\frac{29.43}{2.5}\\\\tan\theta=11.77\\\\\theta=85.14^0

The angle of the initial velocity with respect to the horizontal is 85.14°

4 0
3 years ago
Q.) Miscellaneous conversations. a) mass=120*10^8 g (Convert this value in mg and kg Write in standard form after converting) b)
aleksley [76]

Answer:

a. Convert  120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg

b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm  ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km

Explanation:

a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg

ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg

b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm

ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm  = 2 × 10³ × 10³ cm = 2 × 10⁶ cm

iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km

8 0
3 years ago
during which of the following months is the sun likely to illuminate Earth equally from pole to pole? December January June Sept
Helga [31]

The sun will most likely illuminate Earth equally on both poles during September and March. On these months, Earth will not be tilted toward or away from the sun. This results to both poles receiving nearly the same amount of light, which means that the length of days and nights are equal all over Earth. This event is referred to as Equinoxes. During March, it is called the Vernal Equinox, while during September, it is called the Autumnal Equinox.

 

 

8 0
3 years ago
Read 2 more answers
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