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3 years ago

You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?

Physics

2 answers:

77julia77 [94]3 years ago

5 0

Answer:

20 J of work is done by you, against gravity

Explanation:

Work done by gravity = Potential energy stored in the book after it is put ona shelf.

Potential energy stored:

$U = mgh$

$U = wh$

w= weight in Newton

m= mass of the book

g= acceleration due to gravity

h= height through which, it is raised.

Work is being done against gravity.

$U = 10 \times 2$

$U = 20 J$

Murljashka [212]3 years ago

3 0

Answer:

20 J

Explanation:

Given:

Weight of the book is, $W=10\ N$

Height or displacement of the book is, $d=2\ m$

The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.

So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.

Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.

Therefore, work done is given as:

$Work=W\times d\\Work=10\times 2=20\ J$

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.

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Some bats have specially shaped noses that focus ultrasound echolocation pulses in the forward direction. Why is this useful?

creativ13 [48]

Answer:

The evolutionary success of bats is accredited to their ability, as the only mammals, to fly and navigate in darkness by echolocation, thus filling a niche exploited by few other predators. Over 90% of all bat species use echolocation to localize obstacles in their environment by comparing their own high frequency sound pulses with returning echoes. The ability to localize and identify objects without the use of vision allows bats to forage for airborne nocturnal insects, but also for a diverse range of other food types including motionless perched prey or non-animal food items.

The agility and precision with which bats navigate and forage in total darkness, is in large part due to the accuracy and flexibility of their echolocation system. The echolocation clicks of the few echolocating Pteropodidae (Rousettus) are fundamentally different from the echolocation sounds produced in the larynx that we focus on here, and thus not part of this review. Many studies have shown that bats adapt their echolocation calls to a variety of conditions, changing duration and bandwidth of each call and the rate at which calls are emitted in response to changing perceptual demands . In recent years the intensity and directionality of echolocation signals has received increasing research attention and it is becoming evident that these parameters also play a major role in how bats successfully navigate and forage. To perceive an object in its surroundings, a bat must ensonify the object with enough energy to return an audible echo. Hence, the intensity and duration of the emitted signal act together to determine how far away a bat can echolocate an object. Equally important is signal directionality. Bat echolocation calls are directional, i.e., more call energy is focused in the forward direction than to the sides (Simmons, 1969; Shimozawa et al., 1974; Mogensen and Møhl, 1979; Hartley and Suthers, 1987, 1989; Henze and O'Neill, 1991). An object detectable at 2 m directly in front of the bat may not be detected if it is located at the same distance but off to the side. Consequently, at any given echolocation frequency and duration, it is the combination of signal intensity and signal directionality that defines the search volume, i.e., the volume in space where the bat can detect an object.

The aim of this review is to summarize current knowledge about intensity and directionality of bat echolocation calls, and show how both are adapted to habitat and behavioral context. Finally, we discuss the importance of active motor-control to dynamically adjust both signal intensity and directionality to solve the different tasks faced by echolocating bats.

Explanation:

3 0

3 years ago

Please help <br> Physics is so confusing

crimeas [40]

Answer:

below

Explanation:

sin a = 4/5

a = 53.1

tan theta = 3/4

theta = 36.89

4 0

2 years ago

Starting from rest, a 2-m-long pendulum swings from an angleof

Andrews [41]

Answer:

D.) 1m/s

Explanation:

Assume the initial angle of the swing is 12.8 degree with respect to the vertical. We can calculate the vertical distance from this initial point to the lowest point by first calculate the vertical distance from this point the the pivot point:

$L_1 = L*cos(12.8) = 2*0.975 = 1.95 m$

where L is the pendulum length

The vertical distance from the lowest point to the pivot point $L_2$ is the pendulum length 2m

this means the vertical distance from this initial point to the lowest point is simply:

$L_3 = L_2 - L_1 = 2 - 1.95 = 0.05 m$

As the pendulum travel (vertically) from the initial point to the bottom point, its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the mass of the pendulum, g = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.

The m on both sides of the equation cancel out

$v^2 = 2gh = 2*10*0.05 = 1$

$v = \sqrt{1} = 1 m/s$

so D is the correct answer

5 0

3 years ago

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0

3 years ago

An object with total mass mtotal = 14.6 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 k

zheka24 [161]

Answer: 1) 0. 2) 4.2 Kg. 3) 15.4 m/s 4) 12.9 m/s 5) 0. 6) 3.62 KJ.

Explanation:

1) Assuming that no external forces act during the collision, total momentum must be conserved. As initially the total mass was at rest, so initial momentum is zero, final momentum of all the system must be 0 also.

2) After the explosion, as mass must be conserved also, the sum of the masses of the three pieces must be equal to the original total mass, so we can write the following:

m₁ + m₂ + m₃ = M = 14.6 Kg = 4.9 Kg + 5.5 Kg + m₃

Solving for m₃, we have:

m₃ = 14.6 Kg - 4.9 Kg -5.5 Kg = 4.2 Kg.

3) and 4)

As momentum is a vector, if it is magnitude must be 0, this means that all his components must be 0 too.

So, we can write two equations, one for the x-component, and other for the y-component, as follows:

pₓ = m₁. v₁ₓ + m₂.v₂ₓ + m₃.v₃ₓ = 0

py = m₁.v₁y + m₂. v₂y + m₃. v₃y =0

Replacing by the values, and solving for v₃ₓ and v₃y, we get:

v₃ₓ = 15.4 m/s

v₃y = 12.9 m/s

v = √(15.4)²+(12.9)² = 20.1 m/s

5) As the center of mass must move as if all the mass were concentrated in this point, and we know that the total momentum must be 0, this tells us that the magnitude of the velocity of the center of mass must be 0 too.

6) As initial kinetic energy is 0, as the mass was at rest, the increase in the kinetic energy is obtained simply adding the kinetic energy of every piece of mass gained after explosion, as follows:

K = K₁ + K₂ + K₃ = 1/2 (m₁ . v₁² + m₂.v₂² + m₃.v₃²)

Replacing by the values, we get:

K= 3.62 KJ

4 0

3 years ago

You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?​

You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?