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Vlad [161]
1 year ago
9

Write a conclusion to this lab in which you discuss when a person on a roller coaster ride would have sensations of weightlessne

ss and when they would have sensations of weightiness. In your discussion, talk about accelerations and forces. Then finish off your conclusion by using Newton's second law to explain why such accelerations and force conditions cause these sensations.
Physics
1 answer:
masya89 [10]1 year ago
3 0

Answer:

he lower part of the curve     N = M (g + v² / r)

upper part of the curve          N = m (v² /r -g)

Explanation:

In a roller coaster there is a long climb that allows the car to acquire gravitational potential energy, when this energy is converted into kinetic energy, there is a raven, in these curves we have two parts the lower part, where you have a feeling of great weight and another in the upper part where you have a feeling of weightlessness.

These sensations can be explained using Newton's second law, let's apply it to the lower part of the curve

         N-W = m a

acceleration is centripetal

        a = v² / r

we substitute

        N = mg + m v² / r

        N = M (g + v² / r)

In this part the apparent weight is increased by the speed of the body squared, it feels like a lot of fart.

In the upper part of the curve the force of gravity continues to act downwards, the normal that is the reaction of the surface also goes downwards, the centripetal acceleration pointing towards the center of the curve has a vertical downward direction

        -N -W = -m a

         N = ma -W

         N = m (v² / r -g)

In this case we see that the normal that gives the sensation decreases, which is why we feel a loss of weight, in the case of v2 / r = g, the request is total and the sensation of weightlessness.

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Mr. Chang's class participated in a science experiment where they placed three substances outside in the sunlight to determine w
Montano1993 [528]

Answer:

B or D but im pretty sure it is D

Explanation:

When molecules are left in the sun, it heats up. When molecules heat up, the begin to vibrate rapidly. The sun is not constant as it could get blocked by clouds, so it would, at times, slow down the movement of the molecules. The answer is most likely D.

7 0
2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.
MrRissso [65]
I believe is 10 lb if not it's 9 lb.
5 0
2 years ago
Read 2 more answers
If a ball was thrown upward at 46.3 m/s how long would the ball stay in the air
galben [10]
V = Vo + a.t



The ball is against the vector of gravity. Then, the gravity will be negative.

0 = 46,3  + (-9.8).t \\ 
t =   \frac{46.3}{9.8}  \\ 
t \approx 4.72 



The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.

It will stay approx. 9.44 seconds in the air.
8 0
1 year ago
A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If
sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)=50 \Omega

Voltage(V)=15 V

Time constant(\tau)=\frac{L}{R}

\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}

current i_0=\frac{V}{R}

i_0=\frac{15}{50}=0.3 A

Current as a function of time is given by

i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )

i=0.3\times 0.9998

i= 299.95 mA

6 0
2 years ago
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