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igor_vitrenko [27]
3 years ago
10

A cart of mass M is attached to an ideal spring that can stretch and compress equally well. The cart and spring rest on a smooth

horizontal track. The cart is pulled to position A and released. The cart then moves toward position E, where it reverses direction and returns again to position A.
a) show with the sketch of a position diagram how the velocity of the cart changes between the points indicated
b) The dots below represent the cart at different labeled positions. Draw free body diagrams showing and labeling all the forces exerted on the cart at each labeled position. Draw the relative lengths of all vectors to reflect the relative magnitude of all forces. You will create one set of sketches for when the cart is moving to the right and a second set for when the car is moving to the left.
c) do the diagrams above indicate whether the cart is moving left or right? Justify your claim with evidence.
d) at each position, compare the direction of the net force exerted by the spring on the cart and the carts displacement from equilibrium when at that position. Note that this question is not asking whether the cart is moving right or left. Use these results to briefly explain why your claim in part C makes sense.

Physics
1 answer:
Alexxx [7]3 years ago
4 0

We have to remember a point , which is ' the cart or spring rest on a smooth horizontal track ' , i.e., any frictional force doesn't take place.

<u>Explanation:</u>

a) According to the question the cart is pulled to position A and released, i.e., the velocity of the cart at A initially (say time,t=0) is 0 m/s ,then moves toward position E, where it reverses direction and returns again to position A , in the 2nd phase cart moves along A to E , the cart's velocity increase and again goes to zero at point E and again change the direction, hence

( File has been attached)

b) Let's , the distance between two consecutive points is x meter and the spring constant is k N.m  

c) ( File has been attached)

d) Movinf Right

e) Moving left

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Find the equivalent resistance Req between terminals a and b if terminals c and d are open and again if terminals c and d are sh
adoni [48]

Answer:

351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you  redraw the circuit as below, you can see that the two resistors in the first  column are in parallel as, they are connected together at both pairs of terminals  (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

Now these two sets are  in series with another Hence,

Req = Req1 + Req2 = 216 + 135 = 351 ohms

Answer: 351 ohms

When c and d are shorted:

The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.

Both of these resistor lie in a single path placing the resistors in series to one another, hence

Req = R3 + R1 = 180 + 540 = 720 ohms

Answer:720 ohms

7 0
3 years ago
A dart gun consists of a horizontal spring with k = 52 Newtons/m that is compressed 43
gulaghasi [49]

Answer:

299 m/s^2

Explanation:

When a spring is compressed, the force exerted by the spring is given by:

F=kx

where

k is the spring constant

x is the compression of the spring

In this problem we have:

k = 52 N/m is the spring constant

x = 43 cm = 0.43 m is the compression

Therefore, the force exerted by the spring on the dart is

F=(52)(0.43)=22.4 N

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

F=ma

where

F = 22.4 N is the force exerted on the dart by the spring

m = 75 g = 0.075 kg is the mass of the dart

a is its acceleration

Solving for a,

a=\frac{F}{m}=\frac{22.4}{0.075}=299 m/s^2

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3 years ago
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe
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Answer:

Coefficient of static friction will be equal to 0.642  

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We have given acceleration a=6.3m/sec^2

Acceleration due to gravity g=9.8m/sec^2

We have to find the coefficient of static friction between truck and a cabinet will

We know that acceleration is equal to a=\mu g, here \mu is coefficient of static friction and g is acceleration due to gravity

So \mu =\frac{a}{g}=\frac{6.3}{9.8}=0.642

So coefficient of static friction will be equal to 0.642

3 0
3 years ago
An atom with 4 protons, 5 neutrons, and 4 electrons has an atomic mass of _____ amu. (Enter a whole number.)
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6 0
3 years ago
Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
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