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goldenfox [79]
2 years ago
11

Please follow my sis on i.n.s.t.a.g.r.a.m Id - tehzz_13

Physics
2 answers:
Snezhnost [94]2 years ago
5 0

Answer:

Explanation

ok:

abruzzese [7]2 years ago
4 0

Answer:

btw But I don't use Insta

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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
A man has a mass of 66 kg on earth. What’s his weight in pounds?
Vlad1618 [11]

Answer:

646.8 N

Explanation:

3 0
2 years ago
What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?
vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0
3 years ago
HELP ASAP!!!! WILL GIVE BRAINLIEST!
Pani-rosa [81]

Answer:

Vacculoes, vesicles

Explanation:

Organelles and other structures within the cells have vesicles collectively referred to as the endo-membrane system. The Lysosomes pack up the substances, and the vesicles go through the porsous membrame and secrete the substances.

6 0
2 years ago
a missile is moving 1810 m/s at a 20.0 degree angle. it needs to hit a target 19,500 m away in a 32.0 degree direction in 9.20 s
Ket [755]

Answer:

112 m/s², 79.1°

Explanation:

In the x direction, given:

x₀ = 0 m

x = 19,500 cos 32.0° m

v₀ = 1810 cos 20.0° m/s

t = 9.20 s

Find: a

x = x₀ + v₀ t + ½ at²

19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²

a = 21.01 m/s²

In the y direction, given:

y₀ = 0 m

y = 19,500 sin 32.0° m

v₀ = 1810 sin 20.0° m/s

t = 9.20 s

Find: a

y = y₀ + v₀ t + ½ at²

19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²

a = 109.6 m/s²

The magnitude of the acceleration is:

a² = ax² + ay²

a² = (21.01)² + (109.6)²

a = 112 m/s²

And the direction is:

θ = atan(ay / ax)

θ = atan(109.6 / 21.01)

θ = 79.1°

5 0
2 years ago
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