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Dmitry_Shevchenko [17]

2 years ago

13

A velocity vs. time graph is shown. A graph titled Velocity versus Time shows time in seconds on the x axis, numbered 0 to 5, ve

locity in meters per second on the y axis, numbered 0 to 20. A line starts at the origin and ends at (5, 20). What is the acceleration of the object?

2 answers:

masya89 [10]2 years ago

8 0

Answer:

a+4 m ...

Explanation:

Lisa [10]2 years ago

5 0

Answer:

$a = 4 \frac{m}{s {}^{2} }$

Explanation:

u for velocity

t for time

a for acceleration

$u = a \times t \\ 20 = a \times 5 \\ a = \frac{20}{5} \\ a = 4$

That's it! Hope I helped!

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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

2 years ago

The speed of an object undergoing constant acceleration increased from 8.0 meters per second to 16.0 meters per second in 10. Se

saw5 [17]

v₀ = initial speed of the object = 8 meter/second

v = final speed of the object = 16 meter/second

t = time taken to increase the speed = 10 seconds

d = distance traveled by the object in the given time duration = ?

using the kinematics equation

d = (v + v₀) t/2

inserting the above values in the above equation

d = (16 + 8) (10)/2

d = 120 meter

6 0

2 years ago

Which of the following statements is correct? Which of the following statements is correct? The more a muscle shortens, the more

NeX [460]

The correct answer are

a) "The number of muscle fibers best determines how powerful a muscle will be"

b) "The more a muscle shortens, the more power it generates."

Reason :

Muscle fiber in longitudinal directions generate more power

Multipennate muscles do not produce much power because the tendon branches within muscle .

4 0

2 years ago

A sleeping 68 kg man has a metabolic power of 79 w .

Lesechka [4]

<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>

3 0

2 years ago

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

$h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}$

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV

3 0

2 years ago