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4 years ago

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What advice would a personal trainer give you before a workout?

2 answers:

Katen [24]4 years ago

6 0

Answer:

Have a quick snack an hour before to get the energy, warm up

yulyashka [42]4 years ago

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Answer:

stretch at the end of your workout

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The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

c)

The frequency in a spring-mass system is given by

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

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

#LearnwithBrainly

3 0

3 years ago

To verify the law of solid friction project work of physics

Mars2501 [29]

I am want the anwer to

4 0

2 years ago

What is the mass and volume of 1000kg/m3 of water?

Jobisdone [24]

Answer: The mass would be 1000m3 and the volume would be 1000kg

Explanation:

4 0

3 years ago

Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.

Makovka662 [10]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The Gravitational Force between given objects will be ~

$9.34 \times {10}^{ - 6} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where ~

F = gravitational force

$m_1$ = mass of 1st object = 70 kg

$m_2$ = mass of 2nd object = 2000 kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 1 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 70 \times 2000 }{1 {}^{2} }$

$F =6.674 \times 7 \times 2 \times 10 { }^{ - 11} \times 10 {}^{4}$

$93.436 \times 10 {}^{ - 7}$

$9.3436 \times 10 {}^{ - 6} \: \: newtons$

6 0

3 years ago

Rewire each of the following using the correct prefix using 2 decimal places where applicable.

Ede4ka [16]

Answer:

a. 1.2×10^-6

b. 0.42×10^9

c. 246.8×10^3

d. 88

3 0

3 years ago