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2 years ago

14

Melissa's favorite exercise equipment at the gym consists of various springs. in one exercise, she pulls a handle grip attached

to the free end of a spring to 0.80 m from its unstrained position. the other end of the spring (spring constant = 45 n/m) is held in place by the equipment frame. what is the magnitude of the force that melissa is applying to the handle grip?

1 answer:

konstantin123 [22]2 years ago

6 0

The formula for force exerted on/by a spring is

F = k*e where k is the spring constant and x is the distance stretched from unstrained position. This should allow you to find what you need.

Using F = k x e,

where k is the spring constant,

and e is the extension,

The F is her weight = 45 X 0.80

= 36 N

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A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha

AlexFokin [52]

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as

$m_1u_1+m_2u_2 = (m_1+m_2)v_f$

Here

$m_1$ = Mass of big fish

$m_2$ = Mass of small fish

$v_1$ = Velocity of big fish

$v_2$ = Velocity of small fish

$v_F$ = Final Velocity

The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have

$m_1u_1=-m_2u_2$

$u_2 = -\frac{m_1u_1}{m_2}$

Replacing we have,

$u_2 = -\frac{(5kg)(1m/s)}{0.5kg}$

$u_2 = -10m/s$

The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.

Therefore the speed of the small fish is 10m/s

8 0

2 years ago

Rony fills a bucket with water and whirls it in a vertical circle to demonstrate that the water will not spill from the bucket a

Romashka-Z-Leto [24]

Answer:

0 N, 3.49 m/s

Explanation:

Draw a free body diagram for the bucket at the top of the swing. There are two forces acting on the bucket: weight and tension, both downwards.

If we take the sum of the forces in the radial direction, where towards the center is positive:

∑F = ma

W + T = m v² / r

The higher the velocity that Rony swings the bucket, the more tension there will be. The slowest he can swing it is when the tension is 0.

W = m v² / r

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 1.24 m:

v = √(9.8 m/s² × 1.24 m)

v = 3.49 m/s

8 0

2 years ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

$\texttt{ }$

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<h3>Step B:</h3>

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

Violet light of wavelength 405 nm ejects electrons with a maximum kinetic energy of 0.890 eV from a certain metal. What is the b

AleksandrR [38]

Answer:

Ф = 2.179 eV

Explanation:

This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.

E = K + Ф

the energy of the photons is given by the Planck relation

E = h f

we substitute

h f = K + Ф

Ф= hf - K

the speed of light is related to wavelength and frequency

c = λ f

f = c /λ

Φ = $\frac{hc}{\lambda } - K$

let's reduce the energy to the SI system

K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J

calculate

Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹

Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹

Ф = 3.4571 10⁻¹⁹ J

we reduce to eV

Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

Ф = 2.179 eV

4 0

2 years ago

____ is the force that moves people to behave, think, and feel the way they do, resulting in behavior that is energized, directe

pogonyaev

Answer:

Motivation

Explanation:

Motivation is the force that directs one's behavior. This force is required for repeated actions. There are two types of motivation:

Intrinsic

This type of motivation is comes from the individual

Extrinsic

This type of motivation is comes from an external influence.

Both conscious and unconscious factors influence motivation.

5 0

2 years ago