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Step2247 [10]
3 years ago
9

A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length ℓ and mass m1

extending radially from the surface of a sphere of radius R and mass m2. Find the location of the club's center of mass, measured from the center of the club's head. (Use the following as necessary: m1, m2, ℓ, and R.)
Physics
1 answer:
slega [8]3 years ago
5 0

Answer:

 m₁ (R +ℓ/2) / /( m₁ + m₂ )

Explanation:

We shall consider centre of the club's head  as origin .

The centre of mass of the rod attached with sphere will be away from the centre of sphere by a distance

= R + ℓ /2 and its mass is m₁

mass of sphere is m₂ and distance of its centre of mass from origin is zero

So required CM = m₁ x₁ + m₂ x₂ / ( m₁ + m₂ )

= m₁ (R + ℓ /2) + m₂ x 0 /( m₁ + m₂ )

=  m₁ (R +ℓ/2) / /( m₁ + m₂ )

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A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat
mihalych1998 [28]

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

6 0
3 years ago
PLEASE HELP! I'LL GIVE BRAINLEST​
Harlamova29_29 [7]

Answer:

1.62 m/s²

Explanation:

8 0
3 years ago
A space vehicle is traveling at 2980 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The
Strike441 [17]

Answer:

3054.4 km/h

Explanation:

Using the conservation of momentum

momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor

initial momentum = 14900 M km/h

let v be the new speed of the motor so that the

new momentum = 4Mv and the new momentum of the module  = M ( v + 94 km/h )

total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M

initial momentum = final momentum

14900 M km/h = 5 Mv + 93M

14900 km/h = 5v + 93

14900 - 93 = 5v

v = 2961.4 km/h

the speed of the module = 2961.4 + 93 = 3054.4 km/h

8 0
3 years ago
For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.
raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

8 0
3 years ago
Medical x-rays, with a wavelength of about 10^(-10) m, can penetrate completely through your skin.
natulia [17]
Wavelength of X-rays = 10⁻¹⁰ m
Wavelength of UV = 1000 x 10⁻¹⁰
= 10⁻⁷ m
4 0
3 years ago
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