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MAXImum [283]
3 years ago
12

The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag

ainst the force of gravity?
A) 1.50 × 10^3 J

B) 0.00 J

C) 500 J

D) 4.50 × 10^3 J

I think it is D, but I can also justify B, please help!
Physics
1 answer:
Deffense [45]3 years ago
7 0
The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry
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An example helps clarify the difference between an analysis, a deter- mination and ... departments analyze samples of water to determine the concentration of ... moles of Cu2+, and cylinder 2 contains 20 mL, or 2.0 × 10.
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3 years ago
Using the conservation of mechanical energy, calculate the final speed and the acceleration of a cylindrical object of mass M an
kati45 [8]

Answer:

speed is 4/3 × g × sin β

acceleration is g sin θ

Explanation:

Given data

mass = M

radius = R

distance = s

angle = β

to find out

final speed and the acceleration

solution

we know that cylinder is rolling on inclined plane

so from conservation of mechanical energy

energy at top = energy at bottom

energy at top will be

= m ×g×h

= m ×g×s sinβ    ................1

and

energy at bottom will be

= 1/2 ×m × v² + 1/2 × z × ω²

= 1/2 ×m × v² + 1/2 × 1/2 ×m× r² × (v/r)²

= 3/4 × v²    .................2

equating now equation 1 and 2

m ×g×s sinβ = 3/4 × v²  

so v = 4/3 × g × sin β

so speed is 4/3 × g × sin β

and acceleration is

= mg sin θ    ............3

and

acceleration = ma   ............4

now equation equation 3 and 4

ma = mg sin θ

so acceleration is g sin θ

5 0
2 years ago
A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

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Answer:

Inner Ear

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Psychology is a science because it follows the empirical method

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