What is the control group

2.) Explain why the starting angle doesnt impact the time it takes the pendulum to swing back and forth?

melomori [17]

The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.

The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)

−gsinθ=lθ¨−gain⁡θ=lθ¨

For small angles, θ≪1,θ≪1, and hence sinθ≈θsin⁡θ≈θ. Hence,

θ¨=−glθθ¨=−glθ

This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos⁡(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.

For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause

7 0

3 years ago

A toy mouse has a mass of 4 kg and kinetic energy of 32 J. What is its speed? A. 6 m/s B. 3 m/s C. 5 m/s D. 4 m/s

Svetach [21]

32/4= 8

8/2=4

Answer : D. 4 m/s

7 0

3 years ago

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You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti

vladimir2022 [97]

Answer:

232 J/K

Explanation:

The amount of heat gained by the air = the amount of heat lost by the tea.

q_air = -q_tea

q = -mCΔT

q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)

q = 68,000 J

The change in entropy is:

dS = dQ/T

Since the room temperature is constant (isothermal):

ΔS = ΔQ/T

Plug in values (remember to use absolute temperature):

ΔS = (68,000 J) / (293 K)

ΔS = 232 J/K

5 0

3 years ago

A distracted driver is driving towards a turn where the edge of the road leads into a 75.0 m cliff. The velocity of the vehicle

Vlada [557]

As long as the car is on the road, it moves with a constant speed of 80km/h.

As soon as the car starts to fall down the cliff, it follows a parabolic motion. It means that it still moves with constant speed along the x axis, but it also starts to move along the y axis, with constant acceleration (i.e. the acceleration due to gravity).

The good thing about parabolic motions is that the two motions along the x and y axes are completely separable.

So, first of all, we need to know how long it takes for an object to fall for 75m. The equation of a constantly accelerated motion is

$s=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s_0$ is the initial position, $v_0$ is the initial speed, and $a$ is the constant rate of acceleration. In our case, we start from an initial height of 75m, an initial (vertical!) speed of zero, and our acceleration is $-g$ . So, our equation becomes

$s=75-\dfrac{g}{2}t^2$

And we want to solve for the time when $s=0$ (i.e. we want to know how long will it take for the object to reach the ground). We have

$0=75-\dfrac{g}{2}t^2 \iff 75=\dfrac{g}{2}t^2 \iff \dfrac{2\cdot75}{g}=t^2 \iff t=\sqrt{\dfrac{150}{g}}$

(I'm discarding the negative solution because it wouldn't make sense)

Now that we've used the vertical motion to find out the falling time, we can go back to the horizontal motion. We know that the car moves for a certain amount of time at a certain speed. So, we simply have to plug our values in the $s=vt$ equation, to get

$s=80\sqrt\dfrac{150}{g}}$

This is how far from the base of the cliff the vehicle lands.

3 0

3 years ago

Q6. The 600 N force applied to the bracket at As to be replaced by the two forces, Fa in the a-a direction and Fo in the b-b dir

jeka94

answer:

The horizontal component of the 600 N force is,

Fx = 600 cos(30) 519.61 N

The vertical component of the 600 N force is,

Fy = 600 sin(30) = 300 N

hope it's help!

6 0

3 years ago