A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0

2 years ago

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¿Qué nombre recibe un electrón que abandona el espacio interno del átomo dirigiéndose al exterior del átomo?

Alenkinab [10]

Answer:

i dont speack spanish sorry

Explanation: agian sorry

7 0

2 years ago

Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that th

lisabon 2012 [21]

Answer:

Using equation

qs=F

q=VQ/I V=1200

q=998.45 N

qs=F

s=F/q

s=700/998.45

spacing=s=0.701 m

7 0

2 years ago

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

When you strike two tuning forks simultaneously, you hear three beats per second. The frequency of the first tuning fork is 440

tekilochka [14]

Answer:4

Explanation:

Given

Frequency of beats is 3 beats per second

Frequency of first tuning fork is $f_1=440\ Hz$

Beat frequency is difference in the frequency of tuning forks i.e.

either $f_1-f_2$ or $f_2-f_1 =3$

so $f_2=3+440=443\ Hz$

or

$f_2=440-3=437\ Hz$

so there is not enough information given to decide.

5 0

2 years ago

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A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?​

A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?