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3 years ago

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You have two springs. one has a greater spring constant than the other. you also have two objects, one with a greater mass than

the other. which object should be attached to which spring, so that the resulting spring-object system has the greatest possible period of oscillation?

1 answer:

neonofarm [45]3 years ago

4 0

The answer should be "The object with the greater mass should be attached to the spring with the smaller <span>spring constant."

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

7 0

4 years ago

Read 2 more answers

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

4 years ago

What is another way of saying “getting the smallest possible force”?

Dmitriy789 [7]

Answer:

MOMENTUM

Explanation:

another way of saying getting the smallest force possible is the word " MOMENTUM".

momentum is the ability to keep maintaining,incresing or itself developing to move at constant speed or to increase the speed.

7 0

3 years ago

A 34n force is applied to a 213kg mass how much does the mass accelerate

motikmotik

We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!

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3 years ago

g Which of the following is true about magnetic field lines? A. All magnetic field lines are always parallel to the Earth’s magn

Goshia [24]

Answer:B

Explanation:

Magnetic field lines form close loops and never intercept

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3 years ago