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siniylev [52]
2 years ago
12

The crankshaft in a race car goes from rest to 3000 rpm in 2.0 s.

Physics
2 answers:
Marta_Voda [28]2 years ago
5 0

Answer:

a.  157 rad/s²

b. 50 revolutions

Explanation:

Initial angular velocity, u = 0

Final v = 3000 rpm = 3000 × 2π/60 rad/s= 314 rad/s

time, t = 2.0 s

a. angular acceleration is given by first equation of rotational motion:

α = (v-u)/t = (314 rad/s-0)/ 2.0 s = 157 rad/s²

b. number of revolutions made before reaching final angular velocity of 3000 rpm. Time taken = 2.0 s.

Use second equation of rotational motion:

θ = u t + 0.5 α t²

⇒ θ = 0 + 0.5 × 157 rad/s² × (2.0 s.)² = 314 rad

⇒n ( number of revolutions) =  θ / 2π = 314/ 2π = 50 revolutions

a_sh-v [17]2 years ago
3 0
The equations are analogous to that for linear movement:
acceleration = (final velocity - initial velocity) / time
acceleration = (3000 rpm - 0 rpm) / 2.0 s
a) acceleration = 1500 rpm/s or 25 rp(s^2)
For the displacement
displacement = initial velocity*time + 0.5*acceleration*time^2
displacement = (0)*(2 s) + (0.5)(25 rps^2)*(2 s)^2
b) displacement = 50 revolutions
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3 years ago
Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far
nata0808 [166]

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

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Mass of object 2 = 6.5 x 10³kg

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Unknown:

Distance between them  = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

    Fg  =    \frac{G mass 1 x mass 2}{distance ^{2} }  

   G = 6.67 x 10⁻¹¹

        65  = \frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3}   }{distance^{2} }    

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3 0
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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
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