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2 years ago

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A rock weighing 20 n (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. what is the centr

ipetal acceleration of the rock?

2 answers:

serious [3.7K]2 years ago

8 0

Centripetal acceleration, same as the linear acceleration, is the rate of change of velocity but in this case the tangential velocity. The direction of this acceleration is always directed inward the motion. Centripetal acceleration is calculated from the ratio of the square of the velocity and the radius. We calculate as follows:

centripetal acceleration = v^2 / r
centripetal acceleration = ( 6 m/s )^2 / 2 m
centripetal acceleration = 18 m/s^2

Shkiper50 [21]2 years ago

6 0

Answer:

$a_c = 18 m/s^2$

Explanation:

As we know that the centripetal acceleration is defined as the ratio of square of the speed and the radius of the circle

so it is given as

$a_c = \frac{v^2}{R}$

here we know that

speed of the rock moving in horizontal circle is given as

$v = 6 m/s$

also we know the radius of circle is

$R = 2 m$

now we have

$a_c = \frac{v^2}{R}$

$a_c = \frac{6^2}{2}$

$a_c = 18 m/s^2$

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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

A floating leaf oscillates up and down two complete cycles in one second as a water wave passes by. The wave's wavelength is 10

postnew [5]

Answer:

C) 20 m/s

Explanation:

Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.

The velocity of a moving wave is

v = λf ............................ Equation 1

Where v = speed of the wave, λ = wave length, f = frequency of the wave.

Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters

Substituting these values into equation 1

v = 2×10

v = 20 m/s.

Thus the speed of the wave = 20 m/s

The right option is C) 20 m/s

7 0

2 years ago

A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the hor

Volgvan

Answer:

h=23.67 m : Building height

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t Equation (1)

Where:

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

v₀ = 12.2 m/s , at an angle α=53° above the horizontal

x= 25 m , y=0

Calculation of the time it takes for the ball to hit the ground

We replace data in the equation (1)

x = xi + vx*t

x= 25 m ,xi=0 , vx= v₀*cosα = (12.2 m/s)*cos(53°) =7.34 m/s

25 = 0 + 7.34*t

t= 25 / 7.34

t= 3.406 s

Calculation of the Building height

v₀y = v₀*sinα = (12.2 m/s)*sin(53°) = 9.74 m/s

in t= 3.406 s, y=0

We replace data in the equation (2)

y= y₀ + (v₀y)*t - (1/2)*gt²

0= y₀ + (9.74)*(3.406 )- (1/2)*(9.8)(3.406 )²

0= y₀ + 33.17- -56.84

0= y₀ - 23.67

y₀ = 23.67 m =h: Building height

5 0

2 years ago

A historical society is testing an old cannon. They

sveta [45]

Answer:

26.2 m/s

Explanation:

We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is

$u_y = u sin \theta$ (1)

where u is the initial speed and $\theta = 45.0^{\circ}$ is the angle of projection.

We can therefore use the following suvat equation for the vertical motion of the ball:

$v_y = u_y + at$

where $v_y$ is the vertical velocity at time t, and $a=g=-9.8 m/s^2$ is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:

$t=\frac{3.78}{2}=1.89 s$

And at the maximum height, the vertical velocity is zero:

$v_y=0$

Substituting these values, we find the initial vertical velocity:

$u_y = v_y - at = 0-(-9.8)(1.89)=18.5 m/s$

And using eq.(1) we now find the initial speed:

$u=\frac{u_y}{sin \theta}=\frac{18.5}{sin 45.0^{\circ}}=26.2 m/s$

4 0

2 years ago

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0

2 years ago