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3 years ago

Need help! Photo says all! Will mark brainliest :)

Physics

1 answer:

AVprozaik [17]3 years ago

6 0

Answer:

C-D

Explanation:

As you can see from the graph, the distance from A to B was from 0 m to 6 m in a duration of 3 seconds.

Divide 6 meters by 3 seconds to find the speed:

6 ÷ 3 = 2 m/s

B-C is not moving due to a straight line as said in the graph, so speed is

0 m/s.

There is also C-D since the car traveled from a distance of 9 meters

(6 -(-3) = 9) in 3 seconds too. (NOTE: The graph line going down does not mean it is slowing down, but rather going to a certain distance like going backwards)

Divide 9 meters by 3 seconds to get the speed:

9 ÷ 3 = 3 m/s

Between A-B, B-C, and C-D, C-D has the fastest speed recorded with 3 m/s.

A-D does not count here as the line has no connection between point A and point D.

Cheers!

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Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a

DaniilM [7]

Answer:

a) $k = 1343.6\,\frac{N}{m}$ , b) $l = 0.501\,m\,(50.1\,cm)$

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

$F \propto \Delta l$

In this case, the force is equal to the weight of the object:

$F = m\cdot g$

$F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 80.614\,N$

The spring constant is:

$k = \frac{F}{\Delta l}$

$k = \frac{80.614\,N}{0.408\,m-0.348\,m}$

$k = 1343.6\,\frac{N}{m}$

b) The length of the spring is:

$F = k\cdot (l-l_{o})$

$l = l_{o} + \frac{F}{k}$

$l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }$

$l = 0.501\,m\,(50.1\,cm)$

6 0

3 years ago

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

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3 years ago

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If the motorcycle can go from 0 to 50mph in 3 seconds the person is describing the motorcycle's acceleration.

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