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3 years ago

14

You push against a steamer trunk with a force of 750 N at an angle of 25° with the horizontal . The trunk is on a flat floor and

the coefficient of static friction between the trunk and floor is 0.77. What is the most massive trunk you will be able to move?

1 answer:

djverab [1.8K]3 years ago

3 0

This means that the horizontal force is 750sin(25°). To be able to move the truck, force applied must be greater than static friction, which equals to its coefficient (0.77) x normal contact force (= weight)

Hence, 750sin(25°) = 0.77mg. m = 750sin(25°)/(0.77g)

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The displacement at any given time of an object is x = 6 sin 98t , where the symbols have their usual meanings. i). Proof that t

Fofino [41]

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation, which are represented in the given function as 6 units and 98 rad/s respectively.

<h3>General wave equation for simple harmonic motion</h3>

y = A sinωt

where;

A is amplitude of the motion
ω is angular frequency

<h3>Amplitude of the oscillation</h3>

A = 6 units

<h3>Angular frequency of the wave</h3>

ω = 98 rad/s

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation. Thus, the wave is executing simple harmonic motion.

Learn more about simple harmonic motion here: brainly.com/question/17315536

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2 years ago

3. What is the mass of a paratrooper who experiences an air resistance of 400 N and an acceleration of 4.5 m/s2

goblinko [34]

Answer:

88.89kg

Explanation:

The formula for mass is m=F/a. If we plug in the values, we get m=400N/4.5m/s^2. The mass is 88.89kg. We know that the unit is in kg because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the acceleration, and we are left with kg.

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4 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

a)

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

b)

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

c)

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

d)

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

e)

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

f)

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

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3 years ago

A series RLC circuit has a resistance of 57.61 W, a capacitance of 13.13 mF, and an inductance of 196.03 mH. The circuit is conn

Firdavs [7]

Answer:

voltage across = 1.6 V

Explanation:

given data

resistance R = 57.61 Ω

capacitance c = 13.13 mF = 13.13 × $10^{-3}$ F

inductance L = 196.03 mH = 0.19603 H

fixed rms output Vrms = 23.86 V

to find out

voltage across circuit

solution

we know resonant frequency that is

resonant frequency = 1 / ( 2π√(LC)

put the value

resonant frequency = 1 / ( 2π√(0.19603×13.13 × $10^{-3}$ )

resonant frequency f = 3.1370 HZ

so current will be at this resonant is

current = Vrms / R

current = 23.86 / 57.61

current = 0.4141 A

and

so voltage across will be

voltage across = current / ( 2π f C )

voltage across = 0.4141 / ( 2π ( 3.1370) 13.13 × $10^{-3}$ )

voltage across = 1.6 V

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