(2)

A weightlifter lifts a set of weights a vertical distance of 2 meters. If a constant net force of 350 N is exerted on the weight

o-na [289]

Answer:

The work done by the weightlifter, W = 700 J

The power of the weightlifter, P = 350 watts

Explanation:

A weightlifter lifts a set of weights a vertical distance, s = 2 m

The force exerted to lift the weight, F = 350 N

The work done by the body is defined as the product of the force applied by the body to the displacement it caused.

W = F x s

= 350 N x 2 m

= 700 J

The work done by the weightlifter, W = 700 J

The time taken by the weightlifter to lift the weight, t = 2 s

The power is defined as the rate of body to do work. It is given by the equation,

P = W / t

= 700 J / 2 s

= 350 watts

Hence, the power of the weightlifter, P = 350 watts

6 0

3 years ago

1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential

cupoosta [38]

<h3>Answer:</h3>

$\displaystyle U_s = 0.6 \ J$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: $\displaystyle U_s = \frac{1}{2} k \triangle x^2$

U is energy (in J)
k is spring constant (in N/m)
Δx is displacement from equilibrium (in m)

<h3>Explanation:</h3>

<u>Step 1: Define</u>

k = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

Substitute in variables [Elastic Potential Energy]: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2$
Evaluate exponents: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = 0.6 \ J$

6 0

3 years ago

Spheres of Charge: A metal sphere of radius 10 cm carries an excess charge of +2.0 μC. What is the magnitude of the electric fie

nadezda [96]

To solve this problem we will apply the concept related to the electric field. The magnitude of each electric force with which a pair of determined charges at rest interacts has a relationship directly proportional to the product of the magnitude of both, but inversely proportional to the square of the segment that exists between them. Mathematically can be expressed as,

$E = \frac{kV}{r^2}$

Here,

k = Coulomb's constant

V = Voltage

r = Distance

Replacing we have

$E = \frac{(9*10^9)(2*10^{-6})}{((10+5)*10^{-2})^2}$

$E = 8*10^5N/C$

Therefore the magnitude of the electric field is $8*10^5N/C$

4 0

3 years ago

A little girl kicks a soccer ball, it goes 10 feet then comes back to her, how is this possible??

kirza4 [7]

Its gravitational force

7 0

3 years ago

The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a

umka21 [38]

Let
A = the amplitude of vibration
k = the spring constant
m = the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω = the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
$v_{max} = \omega A$
Therefore
$v(t) = -V_{max} sin(\omega t)$

The KE (kinetic energy) is given by
$KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)$

The PE (potential energy) is given by
$PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)$

When the KE and PE are equal, then
$v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)$

For the oscillating spring,
$\omega ^{2} = \frac{k}{m} \\ V_{max} = \omega A = \sqrt{ \frac{k}{m} } A$
Therefore
$v^{2} = \frac{k}{m} \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max} \,cos( \sqrt{ \frac{k}{m} t} )$

Answer: $v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )$

3 0

4 years ago