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Salsk061 [2.6K]

3 years ago

7

Pin p is constrained to move along the curve defined by the lemniscate r=(4sin2θ)ft.if the slotted arm oa rotates counterclockwi

se with a constant angular velocity of θ˙ = 1.0 rad/s , determine the magnitude of the velocity of peg p when θ = 59 ∘.

1 answer:

ruslelena [56]3 years ago

5 0

position of the peg is given by the equation

$r = 4 sin2\theta$

now the rate of change in position is given as

$v = \frac{dr}{dt}$

$v = \frac{d}{dt}(4 sin2\theta)$

$v = 8cos2\theta*\frac{d\theta}{dt}$

$v = 8 cos2\theta*\omega$

given that

$\omega = 1 rad/s$

$\theta = 59 degree$

now we have

$v = 8*cos(2*59)* 1 = -3.76 m/s$

<em>so its speed will be 3.76 m/s in magnitude</em>

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Nova, whose mass is 50kg drops 2 K, what is the amount of heat lost from Nova's body? (specific heat of the human body is 3470 J

Alex17521 [72]

Use the equation q=mc/\T, where q is the heat lost, m is mass, and /\T is the change in temperature, and c is the specific heat.
q=50kg(3470J/kg K)(2K)
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4 years ago

A vertical piston-cylinder device initially contains 0.1 m^3 of air at 400 K and 100 kPa. At this initial condition, the piston

jenyasd209 [6]

Answer:

$Q=-38.15kJ$

Explanation:

From the question we are told that

Piston-cylinder initial Volume of air $v_1=0.1 m^3$

Piston-cylinder initial temperature $T_1=400k$

Piston-cylinder initial pressure $P_1= 100kpa$

Supply line temperature $T_s=400k$

Supply line pressure $P_s= 500kpa$

Valve final pressure $P_v=500kpa$

Piston movement pressure $P_m=200kpa$

Piston-cylinder final Volume of air $v_2=0.2 m^3$

Piston-cylinder final temperature $T_2=440k$

Piston-cylinder final pressure $P_2= 500kpa$

Generally the equation for conservation of mass is mathematically given by

$Q=m_2 \mu_2-m_1 \mu_1 +W-(m_2-m_1)h$

where

Initial moment

$m_1=\frac{p_1 V_1}{RT_1}$

$m_1=\frac{100*0.1}{0.287*400}$

$m_1=8.7*10^-^2kg$

Final moment

$m_2=\frac{p_2 V_2}{RT_2}$

$m_1=\frac{500*0.3}{0.287*440}$

$m_1=79*10^{-2}kg$

Work done by Piston movement pressure

$W=Pd$

$W=P(v_2-v_1)$

$W=200(0.2-0.1))$

$W=20000J$

Heat function

$h=cT_1$

$h=1.005(400)$

$h=402$

Therefore

$Q=(0.792*0.718(440)-0.0871*0.718(400)+20-(0.792-0.087)*402))$

$Q=-38.15kJ$

It is given mathematically that the system lost or dissipated Heat of

$Q=-38.15kJ$

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3 years ago

What is this object? Explain how it works and which one of Newton’s laws this represents.

lyudmila [28]

Answer:

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4 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

Cart 1 has an initial velocity and hits cart 2 which is stationary. after a perfectly inelastic collision, the combined carts ar

Tomtit [17]

Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.

The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.

Let's demonstrate the issue using variables:

Let,

m1=mass of cart 1

m2=mass of cart 2

v1 = velocity of cart 1 before collision

v2 = velocity of cart 2 before collision

v' = velocity of the carts after collision

Using the conservation of momentum for perfectly inelastic collisions:

m1v1 + m2v2 = (m1 + m2)v'

v2 = 0 because it is stationary

v' = 1/3*v1

m1v1 = (m1+m2)(1/3)(v1)

m1 = 1/3*m1 + 1/3*m2

1/3*m2 = m1 - 1/3*m1

1/3*m2 = 2/3*m1

m2 = 2m1

From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.

To learn more about inelastic collision visit:

brainly.com/question/14521843

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4 0

2 years ago