All categories

4 years ago

When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.

1 answer:

8 0

F=K*X,
F=M*a

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m

You might be interested in

john gottman, the eminent marriage counselor put the number of conflicts which cannot be resolved and needed to be worked on con

USPshnik [31]

He called the counterproductive.

6 0

3 years ago

Which of the following statement is false

just olya [345]

A, the statement is incorrect

6 0

3 years ago

You throw a football straight up. Air resistance can be neglected. (a) When the football is 4.00 m above where it left your hand

Oksana_A [137]

Answer:

The speed of the football when it lift your hand is 8.86 m/s.

Explanation:

Given that,

Final speed of the football, v = 0.5 m/s

Height above the ground, s = 4 m

We need to find the speed of the football when it left your hand. The ball will move under the action of gravity. Using third equation of motion as:

$v^2-u^2=2gs\\\\u^2=v^2-2gs\\\\u^2=(0.5)^2-2\times (-9.8)\times 4\\\\u=8.86\ m/s$

So, the speed of the football when it lift your hand is 8.86 m/s. Hence, this is the required solution.

4 0

4 years ago

An electron, tial well may be anywhere within the interval 2a. So the uncertainty in its position is Î”x= 2a. There must be a co

lapo4ka [179]

Answer:

$K = \frac{h'}{8 m \ \Delta x^2}$ K

Explanation:

The Heisenberg uncertainty principle is

Δx Δp ≥ h' / 2

h’ = $\frac{h}{2\pi }$

The kinetic energy of a particle is

K = ½ m v²

p = mv

v = $\frac{p}{m}$

substitute

K = $\frac{1}{2} \frac{p^2}{m}$

from the uncertainty principle,

Δp = $\frac{h'}{2 \ \Delta x}$

we substitute

K = $\frac{1}{2m} ( \frac{h'}{2 \ \Delta x})^2$

$K = \frac{h'}{8 m \ \Delta x^2}$

4 0

3 years ago

An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.

steposvetlana [31]

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

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

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.

Therefore, T = 0.71 seconds

6 0

3 years ago