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2 years ago

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An oscillating block-spring system has a mechanical energy of 4.72 J, an amplitude of 11.2 cm, and a maximum speed of 4.24 m/s.

Find (a) the spring constant, (b) the mass of the block and (c) the frequency of oscillation.

1 answer:

dsp732 years ago

7 0

Answer:

anyone know this or should i get my brother

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Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

7 0

3 years ago

Read 2 more answers

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

$f'=f(\dfrac{v+v_{p}}{v-v_{s}})$

Where, f = frequency

v = speed of sound

$v_{p}$ = speed of passenger

$v_{s}$ = speed of source

Put the value into the formula

$f'=262\times(\dfrac{344+18}{344-30})$

$f'=302.05\ Hz$

Hence, The frequency is 302.05 Hz.

7 0

3 years ago

What is the length of a spring that has 450J of potential energy and a spring constant of 650N/m?

11111nata11111 [884]

Answer:

Δx = 1.2 m

Explanation:

The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²

Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m

The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.

4 0

3 years ago

the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi

leonid [27]

Answer:24.47

Explanation:

Given

$L_1=30 m$

$d_1=3 mm$

$R_1=25 \Omega$

$L_2=40 m$

$d_2=3.5 mm$

we know Resistance $R=\frac{\rho L}{A}$

Where R=resistance

$\rho =resistivity$

L=Length

A=area of cross-section

$A=\frac{\pi d^2}{4}$

Thus $R\propto \frac{L}{d^2}$

therefore

$R_1\propto \frac{L_1}{d_1^2}$ --------1

$R_2\propto \frac{L_2}{d_2^2}$ --------2

divide 1 and 2 we get

$\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}$

$\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}$

$R_2=25\times 0.734\times 1.33$

$R_2=24.47 \Omega$

8 0

2 years ago

Riley is saving up to buy a new television. She needs a total of $1,298.53. Riley has saved $200.94 already and earns $84.43 per

hoa [83]

$1,298.53 - 200.94 = 1,097.59
$1,097.59 amount still need to save for.
1,097.59/84.43 = 13
13 weeks is the amount of time Riley needs to save enough money for the TV

5 0

3 years ago