Two or more different monomers will link to form a

A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10

Charra [1.4K]

Answer:

$u(t)=1.15 \sin (8.68t)cm$

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

$\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s$

Where $g=980 cm/s^2$

$u(t)=Acos8.68 t+Bsin 8.68t$

u(0)=0

Substitute the value

$A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t$

Substitute u'(0)=10

$8.68B=10$

$B=\frac{10}{8.68}=1.15$

Substitute the values

$u(t)=1.15 \sin (8.68t)cm$

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0

3 years ago

A 562 N trunk is on frictionless plane inclined at 30.0 degrees from the horizontal. What is the acceleration of the trunk down

Len [333]

Answer: 0m/s²

Explanation:

Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane

According newton's law of motion

Summation of forces along the plane = mass × acceleration

Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane

Ff = nR where

n is coefficient of friction = tan(theta)

R is normal reaction = Wcos(theta)

Fm = Wsin(theta)

Substituting in the formula of newton's first law we have;

Fm-Ff = ma

Wsin(theta) - nR = ma

Wsin(theta) - n(Wcos(theta)) = ma... 1

Given

W = 562N, theta = 30°, n = tan30°, m = 56.2kg

Substituting in eqn 1,

562sin30° - tan30°(562cos30°) = 56.2a

281 - 281 = 56.2a

0 = 56.2a

a = 0m/s²

This shows that the trunk is not accelerating

4 0

3 years ago

A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.

Alex73 [517]

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

d' = d cos Ф

= 1430 cos 5.76°

= 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

5 0

3 years ago

A quantity found by multiplying the force by the distance moved

Andrej [43]

The quantity that is calculated from the product of the force and the distance traveled due to the force is called work. It has SI units of Joules (J) which is equivalent to Newton-meter (N-m). It is the energy that happens when an object is being moved by an external force.

7 0

3 years ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago