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pantera1 [17]
2 years ago
11

Argelia has a stack of schoolbooks sitting in the backseat of her car. When Argelia makes a sharp right turn, the books slide to

the left of the seat until they come to a rest against the car door. Explain why this happens.
Physics
2 answers:
mixer [17]2 years ago
8 0

Answer:

Sample Response: Before making the right turn, the books were moving in a straight line. As the turn was being made, the books continued moving in a straight line due to inertia. The books slid across the seat until the car door exerted a force on them that caused them to make the turn with the rest of the car.

Kaylis [27]2 years ago
4 0

Answer:

The books slide to the left due to inertia, and then they come to a rest due to the force applied by the car door.

Explanation:

The entire motion of the books can be explained by using the first two laws of Newton:

- 1st Newton's law (also called Law of Inertia): an object at rest stays at rest and an object in motion stays in motion with constant velocity if no unbalanced forces act on it

- 2nd Newton's law: when an object is acted upon unbalanced forces, an acceleration is induced in the motion of the object, according to the equation

F=ma

where F is the net force on the object, m is its mass and a is its acceleration.

Coming back to our problem:

- At the beginning, Argelia makes a sharp turn right. Due to the law of inertia, the books (which are not fixed to the car) continue their motion straight as it was before the curve: so, since the car is moving right, they appear to go the left of the car.

- When the books hit the car door, they stop moving due to the 2nd Newton's law: in fact, the car door applies an unbalanced force against the books, and as a result the books have a negative acceleration (=deceleration), so they slow down and eventually they stop.

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The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

4 0
2 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

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6 0
2 years ago
At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

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You and your surfing buddy are waiting to catch a wave a few hundred meters off the beach. The waves are conveniently sinusoidal
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Answer:

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T/4= 1.5 s

T= 6 sec

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f= 0.16 Hz (a)

6 0
2 years ago
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