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igor_vitrenko [27]

2 years ago

8

Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

val (on Earth) the Moon clock:
A) (9.8/1.6)h
B) 1 h
C) the square root of 9.8/1.6 h
D) (1.6/9.8)h
E) the square root of 1.6/9.8 h

1 answer:

satela [25.4K]2 years ago

7 0

The time-period of a simple pendulum is

<em>Time = 2 π √(length/grav-accel)</em>

After unraveling the question, then completing it, and working out what I <em>believe</em> it's trying to ask, the choice that correctly answers the question that I have invented is <em>choice-E</em> .

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A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured �� – �� with respect to the positi

Komok [63]

Answer:

223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

$\theta=43^{\circ}$

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a vector= $v_x=vcos\theta$

Where v=Magnitude of vector

Using the formula

x-component of resultant vector= $v_x=8cos43=5.85$

y-component of resultant vector= $v_y=vsin\theta=8sin43=5.46$

x-component of equilibrium vector= $v_x=-5.85$

y-component of equilibrium vector= $-v_y=-5.46$

Because equilibrium vector lies in III quadrant

$\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}$

The angle $\theta'$ lies in III quadrant

In III quadrant ,angle = $\theta'+180^{\circ}$

Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree

4 0

2 years ago

A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an

Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

$\omega = 2\pi f$

$\omega = 2\pi (2.5)$

$\omega = 5\pi rad/s$

The angular displacement is given as the form:

$\theta (t) = \theta_0 cos(\omega t)$

In the equlibrium we have to $t=0, \theta(t) = \theta_0$ and in the given position we have to

$\theta(t) = \theta_0 cos(5\pi t)$

Derived the expression we will have the equivalent to angular velocity

$\frac{d\theta}{dt} = 2.7rad/s$

Replacing,

$\theta_0(sin(5\pi t))5\pi = 2.7$

Finally

$\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°$

Therefore the maximum angular displacement is 9.848°

6 0

2 years ago

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d

Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Speed of the raindrop

- Proposed solution to given ODE:

v = a*t

Where, a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

a = dv / dt ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

v = dx / dt , v = a*t ... proposed solution (Eq 2)

v .dt = dx = a*t . dt

- integrate both sides:

∫a*t . dt = ∫dt

x = 0.5*a*t^2 ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

x (t) = 0.5*a*t^2

x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

m (t) = k*x (t)

m (3.0) = 2.00*x(3.0)

m (3.0) = 2.00*14.7

m (3.0) = 29.4 g

6 0

2 years ago

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri

liubo4ka [24]

Answer:

$T = 7.64 kN$

$F_y = 0.52 kN$ (Downwards)

$F_x = 3.23 kN$ (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

$F(4.50 cos\theta) + mg(2cos\theta) = T \times 3$

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

$5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T$

$T = 7.64 kN$

Now force balance in vertical direction

$F + mg = Tsin65 + F_y$

$5 + 1.40 = 7.64 sin65 + F_y$

$F_y = 0.52 kN$ (Downwards)

Force balance in horizontal direction

$F_x = T cos65$

$F_x = 7.64 cos65$

$F_x = 3.23 kN$ (Towards Left)

7 0

2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers