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alina1380 [7]
3 years ago
14

Use the mass and density data to calculate the volume of corn syrup to the nearest tenth.

Physics
2 answers:
Bas_tet [7]3 years ago
6 0

Answer:

79.1 ml

Explanation:

The density of an object is defined as the ratio of mass of the object to the volume of the object.

Volume of corn syrup = mass of corn syrup x density of corn syrup

Volume of corn syrup = 57.3 x 1.38 = 79.1 ml

Nataly [62]3 years ago
4 0
41.5 is the answer that i got. hope this helps!

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Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
Two particles are moving with positions given by x= 4t^2−2t and x= 6t^3+8t respectively.
Evgen [1.6K]

Explanation:

<em>a)Which of the two has uniform acceleration?</em>

Acceleration is the second derivative of position.  The acceleration of the first particle is:

x = 4t² − 2t

v = 8t − 2

a = 8

The acceleration of the second particle is:

x = 6t³ + 8t

v = 18t² + 8

a = 36t

The first particle has uniform acceleration.

<em>b)Which one is likely to come to rest at some time during its motion?</em>

The particles come to rest when v = 0.  The first particle's velocity has a real zero at t = 4.  The second particle's velocity has only imaginary zeros, meaning v is never 0.

6 0
3 years ago
how do u calculate the kinetic energy of a ball of mass 0.25kg being kicked vertically upwards with a speed of 5m/s​
vfiekz [6]

Answer:

3.125J

Explanation:

K.E.= 1/2(mass)(velocity)^2

K.E.=1/2(0.25)(5)^2=3.125

6 0
3 years ago
❤️❤️❤️❤️WILL MARK BRAINLIEST PLEASE ANSWER BOTH PARTS!❤️❤️❤️❤️
alekssr [168]

Answer:

the answer to the first one is B and the second one is c I hope I got them right

8 0
3 years ago
Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the
kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

U=k\dfrac{q^2}{d}

In this system with three charges which are equidistant from each other

U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}

\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J

The potential energy of the system is 0.12959085 J

6 0
3 years ago
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