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vfiekz [6]
3 years ago
14

In a lever, the effort arm is two times as long as the load arm. The resultant force will be

Physics
2 answers:
Setler [38]3 years ago
7 0
The resultant force will be twice the applied force.

In a lever, the total distance from the applied force to the fulcrum is the effort arm. And, the distance between load and the fulcrum is the load arm.

When the effort arm is two times the load arm, it means the resultant force will be twice the applied force.

Because,  Force = displacement × speed

steposvetlana [31]3 years ago
4 0

The work done by the load arm and the effort arm is the same. If the applied force is f and the length of the effort arm is 2x, the work done by the effort arm is f × 2x. Similarly, the work done by the load arm is F · x. So, F = 2f. Since the effort arm is twice the length of the load arm, the resultant force will be twice the applied force.

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3 years ago
A 200-kg boulder is 1000-m above the ground. what is its potential energy?
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Explanation:

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A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

3 0
3 years ago
An eagle is flying horizontally 16.4 meters above a lake at a speed of 9.3 m/s, carrying a small pumpkin in its talons. The pump
Dima020 [189]

Answer:

The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

Explanation:

Given;

height above the ground, h = 16.4 m

speed of the eagle, v = 9.3 m/s

The time it will take the pumpkin to fall at the given height is calculated as;

t = \sqrt{\frac{2h}{g} }\\\\t =  \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s

The horizontal distance traveled at this time is given by;

x = vt

x = (9.3)(1.83)

x = 17.02 m

Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

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