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Arturiano [62]
3 years ago
5

A bus took 8 hours to travel 639 km. For the first 5 hours, it

Physics
1 answer:
sergey [27]3 years ago
7 0

Answer:

93

Explanation:

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An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl
horsena [70]

Answer:

     A = 2,8333  s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

              T = t/n

calculemos

              A = 34,0/ 12,0

              A = 2,8333  s

6 0
3 years ago
The acceleration due to gravity on the moon is about 5.4ft/s2. if your weight is 150lbf on earth
Ivenika [448]

... then your weight is <em>25.2 lbf</em> on the moon.

6 0
3 years ago
What is the acceleration of a 349 kg object that moved with a force of 750 N?
zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

From the question we have

a =  \frac{750}{349}  \\  = 2.14899713...

We have the final answer as

<h3>2.15 m/s²</h3>

Hope this helps you

4 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du
alisha [4.7K]

Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

Explanation:

The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

ω = Δθ/Δt

ω = \frac{13.9}{5}

ω = 2.78 rad/s

3 0
4 years ago
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
3 years ago
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