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3 years ago

If the spring constant is doubled , what value does the period have for a mass on a spring?

Physics

2 answers:

zhuklara [117]3 years ago

8 0

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

$T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}$

So the correct answer is

D. The period would decrease by sqrt (2)

givi [52]3 years ago

5 0

Answer:

B) the period would be halved by sqrt (2) THIS IS THE RIGHT ANSWER FOR PLATO USERS

Explanation:

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The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in

tangare [24]

The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .

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2 years ago

A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po

juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

$\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}$

$= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}$

$\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}$

$\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\$

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2 years ago

A car is travelling at 15 m/s on a horizontal road and stopped after 4 s. The coefficient of kinetic friction between the tires

givi [52]

Answer:

Fr,= umg

umg= ma

a= v/t

umg= mv/t

u= v/gt= 0.38

3 0

2 years ago

What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

4 0

3 years ago

The net force acting on an object equals the applied force plus the force of friction.

Georgia [21]

Answer:

False

Explanation:

The net force is equal to the applied force minus the force of friction. It is possible for friction to act in the same direction as an applied force, but that would mean there would have to be more than two forces acting on the object.

3 0

2 years ago