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3 years ago

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A 2.0 kg book is lying on a 0.75-m-high table. You pick it up and place it on a bookshelf 2.3 m above the floor. During this pro

cess,

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

The gravitational potential energy gained by the book is 30.4 J

Explanation:

The change in gravitational potential energy of an object is given by

$\Delta U = mg \Delta h$

where

m is the mass of the object

g is the acceleration of gravity

$\Delta h$ is the change in height of the object

For the book in this problem,

m = 2.0 kg

g = 9.8 m/s^2

$\Delta h = 2.3 m - 0.75 m =1.55 m$

Substituting into the equation, we find

$\Delta U = (2.0)(9.8)(1.55)=30.4 J$

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Answer:

because the mass of the copper is higher than the mass of the gold.

Explanation:

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Why do chillies burn even through they’re cold?

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3 years ago

Read 2 more answers

A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen

Sveta_85 [38]

Answer:

$v_x=34 m/s$

$v_y=53.9\ m/s$

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

$v_y=g.t$

The horizontal component of the velocity is always the same:

$v_x=34 m/s$

The vertical component at t=5.5 s is:

$v_y=9.8*5.5=53.9$

$v_y=53.9\ m/s$

8 0

3 years ago

How many cycles have been completed after 3 seconds on the graph

Reptile [31]

Answer:

3 Cycles

Explanation:

Every cycle on a cosine function are marked by the completion of 5 key points when moving along the x-axis:

(5, 0, -5, 0, 5)

This cycle has a frequency of 2, occurring once every second, thus answering our question easily.

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3 years ago

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Juliette [100K]

<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, <u>as long as the spring is not permanently deformed</u>:

$F=k (x-x_{o})$ (1)

Where:

$k$ is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

$x_{o}$ is the length of the spring without applying force.

$x$ is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth $W$ of the block:

$W=m.g$ (2)

Where $m=250g=0.25kg$ is the mass of the block and $g=9.8\frac{m}{s^{2}}$ is the gravity acceleration.

$W=(0.25kg)(9.8\frac{m}{s^{2}})$ (3)

$W=2.45N$ (4)

Knowing the force applied $W$ and $x=18cm=0.18m$ and $x_{o}=0$ , we can substitute the values in equation (1) and find $k$ :

$W=k (x-x_{o})$ (5)

$2.45N=k (0.18m-0m)$ (6)

<u>Finally:</u>

$k=13.61\frac{N}{m}$

4 0

3 years ago