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telo118 [61]
3 years ago
12

You are at the park with your little brother, when you notice a small merry-go-round with a radius that looks to be about 1.5 m.

Your brother climbs on, and you give him a spin. From rest, you speed up smoothly, completing one full lap in 3.0 seconds. You wonder what sort of acceleration (magnitude and direction) your brother experiences at the very start of the motion, and at the very end of the first lap. He is quite young and hasn't studied any physics, so he does not know how to answer your question. You decide to give him your smartphone (which has a built-in accelerometer), and repeat the experiment. What will the smartphone record
Physics
1 answer:
Elodia [21]3 years ago
7 0

Answer:

Explanation:

angle covered in one rotation = 2π radian

θ = ωt + 1/2 αt²

θ is angle rotated in time t with initial angular velocity of ω and angular acceleration α .

Putting the values

2π = 0 + 1/2 x α x 3²

α = 1. 4 radian / s²

linear acceleration =  α x r = 1.4 x 1.5 = 2.1 m / s².

Initial acceleration = 2.1 m /s²

final angular velocity = α t = 1.4 x 3 = 4.2 radian / s

linear velocity = 4.2 x 1.5 = 6.3 m /s

centripetal acceleration = v² / R = 6.3² / 1.5 = 26.46 m /s²

radian acceleration = 26.46 m /s

tangential acceleration = 2.1 m /s²

Total final acceleration = √ ( 26.46² + 2.1² )

= √ ( 700.13 + 4.41)

Final acceleration = 26.53 m / s²

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Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring
lana [24]

Answer:

y (t) = 0.754 * cos ( 7.96 t - 69.52)

Explanation:

Given:

m = 1.5 kg , k = 95 N / m , v₀ = 6 m / s , d = 0.35 m , t = 0

y (t) = A * cos ( ω * t - φ )

Using the equation that describe the motion

m * v = - k * x  ⇒ m * x'' = - k * x

Angular velocity is equal to

ω = √ k / m   ⇒ ω = √ 95 N /m  /  1.5 kg  

ω = 7.96  rad /s

A = v / ω   ⇒  A = 6 m /s   /   7.96 rad / s

A = 0.754

d = cos * φ    ⇒  φ = cos ⁻¹ * 0.35

φ = 69.52

y (t) = A * cos ( ω * t - φ )    ⇒  y (t) = 0.754 * cos ( 7.96 t - 69.52)

4 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
What type of physical weathering is most common in a river
tia_tia [17]

Answer:

Exfoliation or Unloading. As upper rock portions erode, underlying rocks expand. ...

Thermal Expansion. Repeated heating and cooling of some rock types can cause rocks to stress and break, resulting in weathering and erosion. ...

Organic Activity. ...

Frost Wedging. ...

Crystal Growth.

Explanation:

3 0
4 years ago
Read 2 more answers
Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc
Tasya [4]

Answer:

The rock's speed after 5 seconds is 98 m/s.

Explanation:

A rock is dropped off a cliff.

It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².

Speed after 5 seconds = V

We know that acceleration = average speed/time

In our case,

g = ((0+V)/2)/5

9.8*5 = V/2

=> V = 2*9.8*5

V = 98 m/s

3 0
3 years ago
Read 2 more answers
A fidget spinner experiences a constant torque of 1.4 N.m. If the spinner is initially at rest, what is its angular momentum 2.0
motikmotik

Answer:

2.8N-msec

Explanation:

We have given torque \tau =1.4N-m

Initial time t_1=0\ sec

Final time t_2=2\ sec

There is relation between angular momentum and torque

that is \frac{dL}{dt}=\tau

dL=\tau dt

\int dL=\tau \oint_{t_1}^{t_2}dt

L=\tau (t_1-t_2)=1.4\times (2-0)=2.8N-msec

4 0
4 years ago
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