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andre [41]
4 years ago
6

What is an object’s acceleration if it is moving at 30 m/s and comes to a stop in 5 s? –30 m/s2 –6 m/s2 30 m/s2 6 m/s2

Physics
2 answers:
Readme [11.4K]4 years ago
7 0

Answer:

The acceleration is  -6 m/s2

Explanation:

The equation for the speed of a body is

         V = Vo + a t

Where

Vo is the initial velocity, a is the acceleration and t is the time,

When the body stops the speed is zero

       0 = Vo + a t

        a = Vo / t

substitute values

        a = - 30/5

        a = - 6 m/s2

The negative sign indicates that acceleration opposite movement

Tanya [424]4 years ago
3 0

Answer:32m/s

Explanation:

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The mass of four washers is  0.0196 kg.

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What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the
timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B =  $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $

Therefore, the maximum initial K.E. = 3066.67 J

3 0
3 years ago
Summarize the big bang theory
MrRissso [65]
Boom explosion then your parents made u :$$
8 0
3 years ago
Please Help No Link Seriously
Artist 52 [7]

Answer:

x = 8.3 cm

y = 5.4 cm

Explanation:

x position

200(0) + 100(12) + 350(12) / (200 + 100 + 350) = 8.307692...

y position

200(0) + 100(0) + 350(10) / (200 + 100 + 350) = 5.3846153...

7 0
3 years ago
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