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VladimirAG [237]
3 years ago
5

Hi, I was wondering why do in real life we refer our weight as for example 54kg instead of 540N? When we know that weight is a f

orce and the S.I unit is Newton.
Physics
1 answer:
Kazeer [188]3 years ago
8 0

Answer:

Chicken nuggets

Explanation:

Because

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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
You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,
telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

t_{1}=2.19\ sec

t_{2}=2.30\ sec

t_{3}=2.26\ sec

t_{4}=2.29\ sec

t_{5}=2.27\ sec

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

t_{1}=2.19\approx 2.2\ sec

t_{2}=2.30\approx 2.3\ sec

t_{3}=2.26\approx 2.3\ sec

t_{4}=2.29\approx 2.3\ sec

t_{5}=2.27\approx 2.3\ sec

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0
3 years ago
A 0.15 kg project hits the ground with a speed of 11 m/s. The project comes to rest in 0.015 seconds. What is the net force?
Blababa [14]

Answer:

Explanation:

ACCORDING TO NEWTONS SECOND LAW;

F = mass * acceleration

F = m(v-u/t)

m is the mass = 0.15kg

v is the final velocity = 11m/s

u is the initial velocity = 0m/s

t is the time = 0.015

Substitute;

F = 0.15(11-0)/0.015

F = 0.15(11)/0.015

F = 1.65/0.015

F = 110N

Hence the net force is 110N

4 0
2 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
Students hypothesized that by running an electric current through the wire of the apparatus shown here, they could cause a non-m
astra-53 [7]
The answer is d i think
5 0
3 years ago
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