Before beginning an exercise program a person should always

The following four waves are sent along strings with the same

likoan [24]

Answer:

Explanation:

y_1 = (3 mm) sin(x - 3t)

comparing it with standard wave equation

y = A sin( ωt-kx )

we see

ω = -3 , k = -1

velocity = ω / k

= 3

y_2 = (6 mm) sin(2x - t)

we see

ω = -1 , k = -2

velocity = ω / k

= .5

y_3 = (1 mm) sin(4x - t)

we see

ω = -1 , k = -4

velocity = ω / k

= .25

y_4 = (2 mm) sin(x - 2t)

we see

ω = -2 , k = -1

velocity = ω / k

= 2

So greatest velocity to lowest velocity

y_1 = (3 mm) sin(x - 3t) , y_4 = (2 mm) sin(x - 2t) ,y_2 = (6 mm) sin(2x - t) , y_3 = (1 mm) sin(4x - t)

b )

Given the mass per unit length of wire the same , velocity is proportional to

√ T , where T is tension

so in respect of tension in the wire same order will exist for highest to lowest tension .

3 0

3 years ago

A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m

katovenus [111]

Answer:

217.43298 m/s

Explanation:

$m_1$ = Mass of bullet = 19 g

$m_2$ = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

$\theta$ = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s$

As the momentum is conserved

$m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s$

The speed of the bullet is 217.43298 m/s

5 0

4 years ago

The effect of gravity is more in liquid than in solid. why?

ycow [4]

This is not that exact but I hope that this will help you.

The effect of gravity is more in liquid than in solid because as we know that liquid pressure increase in the increase in depth but solid dont

5 0

4 years ago

Read 2 more answers

How to solve part (b)?

maria [59]

Answer:

1.11m

Explanation:

From the diagram we are given the following forces;

F1 = 24.3N

F3 = 30N

Since the sum of upward forces is equal to that of downward force, then;

F2 = F1 + F3

F2 = 24.3N + 30N

F2 = 54.3N

Required

Distance between B and C

First we need to get Length of AC

Take moment about A

Anticlockwise moment = F3 cos20 * AC

Anticlockwise moment = 30ACcos 20

Clockwise moment = 1.2 * F2

Clockwise moment = 1.2(54.3) = 65.16Nm

Applying the principle of moment;

Sum of ACW moment = Sum of CW moments

30ACcos 20 = 65.16

AC = 65.16/30cos20

AC = 65.16/28.19

AC = 2.31m

Get the distance BC

AC = AB + BC

BC = AC-AB

BC = 2.31 - 1.2

BC = 1.11m

Hence the separation between B and C is 1.11m

Note that the force F1 got in (a) was the value used in the calculation.

8 0

3 years ago

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

3 years ago