The color that an object appears to be depends on the select one:

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time inte

ddd [48]

We can answer this using the rotational version of the kinematic equations:
θ = θ₀ + ω₀t + ½αt² -----> 1

ω² = ω₀² + 2αθ -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² 

ω₀ = (70.4 + 37.544) / 3.80 

ω₀ = 28.406 rad/s 

Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4

ω = 8.65 rad/s

5 0

3 years ago

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

34kurt

Answer:

The centripetal acceleration of the runner is $1.73\ m/s^2$ .

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

v is the velocity of runner

$v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s$

Centripetal acceleration,

$a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2$

So, the centripetal acceleration of the runner is $1.73\ m/s^2$ . Hence, this is the required solution.

5 0

3 years ago

A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and

vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

$U=mgh$

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

$h=4-4cos(30\°)=0.53m$

Then, the potential energy is:

$U=(55kg)(9.8m/s^2)(0.53m)=285.67J$

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

$U'=(m_1+m_2)gh'=285.67\ J$

From the previous equation you can get h':

$h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m$

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

$h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°$

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0

3 years ago

A parallel plate capacitor filled with air with plate area 2-cm2 and plate separation of 0.5 mm is connected to a 12 V batter an

pashok25 [27]

Answer:

a) The charge of the capacitor is 4.25x10⁻¹¹C

b) The charge of the capacitor is 4.25x10⁻¹¹C because the battery is disconnected.

c) The potential difference across the plates is 18 V

d) The work is 7.64x10⁻¹⁰J

Explanation:

The capacitance of the capacitor is equal to:

$C=\frac{e_{0}A }{d}$

A = 2 cm² = 0.0002 m²

d = 0.5 mm = 0.0005 m

Replacing:

$C=\frac{8.85x10^{-12}*0.0002 }{0.0005} =3.54x10^{-12} F = 3.54pF$

a) The charge of the capacitor is equal to:

Q = C*V = 3.54 * 12 = 42.48 pC = 4.25x10⁻¹¹C

b) The charge is the same because the battery is disconnected (Q = 4.25x10⁻¹¹C)

c) If distance is increased, we have:

$C=\frac{8.85x10^{-12}*0.0002 }{0.00075} =2.36x10^{-12} F=2.36pF$

The potential is:

$V=\frac{Q}{C} =\frac{42.48}{2.36} =18V$

d) The work done is equal to:

$W=VQ=18*42.48=764.6pJ=7.64x10^{-10} J$

3 0

3 years ago

Read 2 more answers

The quantity of charge through a conductor is modeled as Q = (3.00 mC/s4)t4 − (2.00 mC/s)t + 9.00 mC. What is the current (in A)

Mumz [18]

Answer:

The current at time t = 4.00 s is 0.766 A.

Explanation:

Given that,

The quantity of charge through a conductor is modeled as :

$Q=(3t^4-2t+9)\ mC$

We need to find the current (in A) at time t = 4.00 s. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(3t^4-2t+9)}{dt}\\\\I=12t^3-2$

At t = 4 s

$I=12(4)^3-2=766\ mC/s\\\\I=0.766\ C/s=0.766\ A$

So, the current at time t = 4.00 s is 0.766 A. Hence, this is the required solution.

7 0

3 years ago