50 points for this please help

A blender takes in 180 J of electrical energy. Of the 180 J, 90 J are transformed into kinetic energy, 60 J are transformed into

a_sh-v [17]

Fa sho it's 30j edge said so

6 0

3 years ago

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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

A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow

valentina_108 [34]

F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer

7 0

3 years ago

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Example of first, second and third law of motion.

statuscvo [17]

Examples of Newton's three law of motion.

First law of motion: A rocket being launched up in the atmosphere.

Second law of motion:while riding a bicycle, a bicycle acts as a mass and our legs pushing on the pedals of the bicycle is the force.

Third law of motion:when we jump off from the boat,the boat moves backward.

Hope,it will helpyouu!

4 0

2 years ago

a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec

MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

$\frac{1}{2} mu_1^2+0=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $\frac{1}{2} m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2$

∴ $\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $mv$ ₁ $v$ ₂ = 0

It is impossible for the mass to be zero.
Because the second ball moves, velocity v2 cannot be zero.
As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.

<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0

1 year ago