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2 years ago

it is possible that the acceleration and velocity are perpendicular to each other? explain with example

Physics

1 answer:

shusha [124]2 years ago

5 0

Answer: Ok so We already know that velocity is on the x-axis.

Since acceleration = Force / Mass

Here the Force is downward due to the gravitational pull or we can say it is along y-axis.

Since acceleration is directly proportional to force, so acceleration is also along y-axis. This means that velocity & acceleration are perpendicular to each other.

Example:

Let us assume that an aeroplane is flying parallel to the horizontal plane. The aeroplane will experience the acceleration in several directions. One of them here is the gravitational pull which is perpendicular to the the apparent velocity. So the net velocity & its direction will depend upon the vector sum total of all the forces/acceleration acting on it. Also because of this gravitational pull the aeroplane rotates along with the earth, which is a proof that the force/g experienced by it does not go waste.

<h3>Hope this helps have a awesome day/night❤️✨</h3>

Explanation:

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Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t

ruslelena [56]

Answer:

$x_1= 0.0425m$

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

$m*g = k*x$

With two springs, let, T1 be the tension in each spring, x1 be the extension of each spring. The spring constant of each spring is 2k so:

$T_1 = 2k*x_1$

$2T_1 = m*g=4k x_1$

Solve to x1

$x_1=\frac{m*g}{4k}$

$x_1=\frac{k*x}{4*k}$

$x_1=\frac{x}{4}$

$x_1 = 0.170 / 4$

$x_1= 0.0425m$

7 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:

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:

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

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:

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

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.

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

is the following sentence true or false? the faster the particles of a substance are moving, the more energy they have.

svetlana [45]

I think true. I'm pretty sure, but check w/ others too.

8 0

3 years ago

Read 2 more answers

Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el

kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

$E =\frac{Kq}{r^2}$

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

$q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9}$ = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

8 0

3 years ago

A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict

lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

$\sum F = ma$