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Butoxors [25]
4 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]4 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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mezya [45]

Answer:

No,it isn't concave. The correct answer is convex lens.

Explanation:

<em>A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.</em>

7 0
3 years ago
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Which is an example of the force of attraction between two objects that have mass?
Genrish500 [490]

According to multiple scientific experiments, most objects develop an attraction towards Candice.

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3 years ago
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HELPPP ASAP!!!!!!
fomenos

Answer:

The acceleration of the player is - 4.9 m/s²

Explanation:

The given is:

1. The mass of the player is 55 kg

2. His initial speed is 4.6 m/s

3. The coefficient of the kinetic fraction between the player and the

    ground is 0.50

We need to find the player acceleration

According to Newton's Law

→ ∑ forces in direction of motion = mass × acceleration

There is only the friction force opposite to the motion

→ Friction force = μR

where μ is the coefficient of friction and R is the normal reaction

→ The normal reaction R = mg

where m is the mass and g is the acceleration of gravity

→ m = 55 kg , g = 9.8 m/s²

→ R = 55 × 9.8 = 539 N

→ ∑ F = - μR

→ - μR = m × a

→ μ = 0.5 , R = 539 N , m = 55

→ -(0.5)(539) = 55 × a

→ - 269.5 = 55 a

Divide both sides by 55

→ a = - 4.9 m/s²

The acceleration of the player is - 4.9 m/s²

Learn more:

You can learn more about Newton's law in brainly.com/question/11911194

#LearnwithBrainly

3 0
4 years ago
A particle of mass m moves under the influence of a force given by F = (−kx + kx3/α2) where k and α are positive constants. a) F
Romashka-Z-Leto [24]

Answer:

Explanation:

a ) F = (-kx + kx³/a²)

intensity of field

I = F / m

=  (-kx + kx³/a²) / m

If U be potential function

- dU / dx =  (-kx + kx³/a²) / m

U(x)  = ∫  (kx - kx³/a²) / m dx

= k/m ( x²/2 - x⁴/4a²)

b )

For equilibrium points , U is either maximum or minimum .

dU / dx = x - 4x³/4a² = 0

x = ± a.

dU / dx = x - x³/a²

Again differentiating

d²U / dx² = 1 - 3x² / a²

Put the value of x = ± a.

we get

d²U / dx²  = -2 ( negative )

So at x = ± a , potential energy U is maximum.

c )

U =  k/m ( x²/2 - x⁴/4a²)

When x =0 , U = 0

When x=  ± a.

U is maximum

So the shape of the U-x curve is like a bowl centered at x = 0

d ) Maximum potential energy

put x = a or -a in

U(max)  =  k/m ( x²/2 - x⁴/4a²)

= k/m ( a² / 2 - a⁴/4a²)

= k/m ( a² / 2 - a²/4)

a²k / 4m

This is the maximum total energy where kinetic energy is zero.

4 0
4 years ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
SSSSS [86.1K]

Answer:

{F_{tot} = -1.092*10^{-2}N

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

F =k\dfrac{q_1 q_2}{d^2}.

where d is the separation between the charges q_1 and q_2.

Now, in our case

q_1=-13.5nC=-13.5*10^{-9}C

q_2=-1.735nC=-1.735*10^{-9}C

q_3= 47*10^{-9}C

The separation between charges q_3 and q_1 is

d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m

Therefore, the force between them is

F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N

and it is directed in the negative x-direction.

The separation between charges q_2  and q_3 is

d_{23} =-1.240mm-0=-1.240*10^{-3}m

therefore, the force between them is

F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N

Therefore the total force on charge q_3 is

F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}

8 0
3 years ago
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