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Sav [38]
3 years ago
13

Rafael swings a baseball bat with a force of 25 N. The baseball bat exerts a force of 5 N on the baseball. What is the mechanica

l advantage of the baseball bat? The mechanical advantage of the baseball bat is
Physics
2 answers:
goldfiish [28.3K]3 years ago
7 0

Answer:

0.2

Explanation:

Mechanical Advantage is defined as the ratio of force output with respect to the force input. Thus,

MA = output / input

MA = 5 / 25

MA = 1/5

MA = 0.2

Since this value is less then one. Even though it's tagged a mechanical advantage, in reality it would be mechanical disadvantage.

Therefore, we can say that the mechanical advantage of the baseball bat is 0.2

Firdavs [7]3 years ago
5 0

Answer:

0.2

Explanation:

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A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel
lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: -10 [N] = m.ax

Y-direction balance force: -m*9,8 \frac{m}{s^2} = m.ay

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

ax=(-10 [N]) / m

ay=-m*9,8\frac{m}{s^2} / m

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

ay=-9,8\frac{m}{s^2}

For the x component aceleration we must replace the Newton unit:

N =\frac{kg.m}{s^2}

ax= -10 \frac{kg.m}{s^2} / (2 kg)

ax= - 5 \frac{m}{s^2}

6 0
3 years ago
A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?
Triss [41]

Answer:

The angle of projection is 12.26⁰.

Explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

H_m = \frac{u^2sin^2 \theta}{2g}

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta  = sin^{-1}(0.2123)\\\\\theta  = 12.26^0

Therefore, the angle of projection is 12.26⁰.

6 0
2 years ago
A 783kg elevator rises straight up 164 meters. What is the potential energy of the elevator?
Vilka [71]

Answer:

potential \: energy = mgh \\ m = 783 \\ g = 10 \\ h = 164 \\ pe = 783 \times 10 \times 164 \\ =  7830 \times 164 \\  = 1284120 \: joule \\ thank \: you

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