Answer:
x = 12cm
Explanation:
First, with the values given for the force and the displacement, you calculate the spring-constant of the dynamometer by using the following equation (primero, con los valores dados de la fuerza y el desplazamiento, calculas la constante del resorte aplicando la siguiente formula):
F: force (fuerza) = 10N
k: spring constant (constante del resorte)
x: displacement of the spring (desplazamiento del resorte) = 6cm
by using these values you calculate k (usando estos valores calculas k):
With this value you calculate the displacement for a force of 20N (con este valor de k calculas el desplazamiento del resorte para una fuerza de 20N):
hence, the displacement is 12cm
Answer:
H₀ = 1.6 x 10⁻¹⁸ s⁻¹
Explanation:
The Hubble's Constant can be found by the following formula:
where,
H₀ = Hubble's Constant = ?
v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s
D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)
D = 18.52 x 10²⁴ m
Therefore,
<u>H₀ = 1.6 x 10⁻¹⁸ s⁻¹</u>
Answer:
The speed will be "18km/s". A further explanation is given below.
Explanation:
According to the question, the values are:
Wavelength,
As we know,
⇒
On substituting the values, we get
⇒
⇒
⇒
⇒
or,
⇒
Answer:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
Explanation:
The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>
We have a block-spring system, whose angular frequency () is defined by the following formula:
(1)
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
And the period (), measured in seconds, is determined by the following expression:
(2)
By applying (1) in (2), we get the following formula:
a) If the mass is doubled, then the period is increased by . Hence, the period of the system is 2.828 seconds.
b) If the mass is halved, then the period is reduced by . Hence, the period of the system is 1.414 seconds.
c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.
d) If the spring constant is doubled, then the period is reduced by . Hence, the period of the system is 1.414 seconds.