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3 years ago

7

Given the units of force, write a simple equation relating a constant force Fexerted on an object, an interval of time t during

which the force is applied, andthe resulting momentum of the object,

1 answer:

sattari [20]3 years ago

3 0

newton's second law ...

F=momentum change/time

F = (Final mom'm-start mom'm)/time

Ft+start =final

Ft is impulse

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In a traveling or following spotlight in a musical theater production, the brilliant xenon lamp positioned

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The answer is the letter a

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Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th

Anit [1.1K]

Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components. While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

4 0

3 years ago

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0

3 years ago

What momentum does a 70 kg person running 10m/s (a fast sprint) have ?

gogolik [260]

Momentum = (mass) x (speed)

Momentum = (70 kg) x (10 m/s)

<em>Momentum = 700 kg-m/s</em>

8 0

3 years ago

A 12.0-g plastic ball is dropped from a height of 2.50 m. Just as it strikes the floor, it is moving at a speed of 3.20 m/s. How

nalin [4]

Answer:

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

6 0

3 years ago