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3 years ago

Big Bang theory flow chart

Physics

1 answer:

swat323 years ago

6 0

Answer:

I don't understand

can please say it again

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Using the knowledge you gained from your lessons and from this practical exercise, this engine appears to be a

Travka [436]

D)
(()(){({)({)(()({)())({)()({)()()()9

4 0

3 years ago

UNO [17]

Calculate the minimum speed record at the point B in order for the real question to reach the top

7 0

3 years ago

How long must a simple pendulum be if it is to make exactly one swing per four seconds? (That is, one complete oscillation takes

cricket20 [7]

Answer:

Length of the pendulum will be 3.987 m

Explanation:

We have given time period of the pendulum T = 8 sec

Acceleration due to gravity $g=9.81m/sec^2$

We have to find the length of the simple pendulum

We know that time period of the simple pendulum is given by

$T=2\pi \sqrt{\frac{l}{g}}$

$8=2\times 3.14 \sqrt{\frac{l}{9.81}}$

$l=3.987m$

So length of the pendulum will be 3.987 m

3 0

3 years ago

A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2

madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0

3 years ago

The coefficient of friction between a rider and the merry go round is 0.45 and the person is measured to be traveling at 20.0 RP

svlad2 [7]

The person must stand at a radius of 0.99 m

Explanation:

In order for the person to stand on the merry go round, the force of friction acting on the person must provide the centripetal force necessary to keep the person in uniform circular motion.

Therefore, we can write:

$\mu mg = m\omega^2 r$

where:

- the term on the left is the force of friction, and the term on the right is the centripetal force

$\mu = 0.45$ is the coefficient of friction

m is the mass of the person

$g=9.8 m/s^2$ is the acceleration of gravity

$\omega = 20 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=2.09 rad/s$ is the angular velocity

r is the radius of the circular path

Solving the equation for r, we find the radius at which the person must be standing:

$r=\frac{\omega^2}{\mu g}=\frac{(2.09)^2}{(0.45)(9.8)}=0.99 m$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago