All categories

1 year ago

Please help with this question, step by stepThank you!

Physics

1 answer:

mamaluj [8]1 year ago

7 0

From the calculation, the force constant is 192 N. Also, friction would decrease the extension.

<h3>What is the force constant?</h3>

We know that the force constant can be obtained by the use of the relation;

F = Ke

F = applied force

K = force constant

e = extension

We know from Hooks law that the force applied is directly proportional to the extension.

We can write;

F = mgcosθ

F = 43 Kg * 9.8 m/s^2 * sin31°

F = 217 N

K = 217 N/1.13 m

K = 192 N/m

If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.

Learn more about Hooke's law:brainly.com/question/14140269

#SPJ1

You might be interested in

A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr

Musya8 [376]

Answer:

162500000.

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

$n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}$

n=162500000 t

$\dfrac{n}{t}=16250000$

The number of electron passe per second will be 162500000.

4 0

3 years ago

What type of diagram is this?

FromTheMoon [43]

Answer: That’s a sankey diagram

Explanation:

5 0

3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

3 years ago

Can a runner on a track be accelerating even if she is running at a constant speed why or why not

Lynna [10]

If the runner is running in a circular track then yes when something or someone is moving in a circular motion at a constant speed they are indeed accelerating. They’re accelerating because the direction of the velocity vector is changing

6 0

3 years ago

A 11.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1985 J by the time the rock

nirvana33 [79]

Answer:

h = 18.41 m

Explanation:

Given that,

Mass of a test rocket, m = 11 kg

Its fuel gives it a kinetic energy of 1985 J by the time the rocket engine burns all of the fuel.

According to the law of conservation of energy,

PE = KE = mgh

h is height will the rocket rise

$h=\dfrac{E}{mg}\\\\h=\dfrac{1985 }{11\times 9.8}\\\\h=18.41\ m$

So, the rocket will rise to a height of 18.41 m.

5 0

3 years ago