Question 5 of 10

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

4 years ago

A car accelerates for 10 seconds. During this time, the angular

bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, $\omega_i$ = 10 rad/s

final angular velocity, $\omega_f$ = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

$a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2$

Therefore, the angular acceleration of the car is 1.5 rad/s²

4 0

3 years ago

Which of these atoms is the most electronegative? select one: a. si b. cl c. p d. f e. c

katrin [286]

The atom that is the most electronegative is fluorine (F).

<h3>What is electronegative?</h3>

Electronegativity, is the tendency for an atom of a given chemical element to attract shared electrons when forming a chemical bond.

Electronegativity increases across the groups from left to right of the periodic table and decreases down the group.

Examples of electronegative elements arranged in decreasing order;

fluorine,
oxygen,
nitrogen,
chlorine,
bromine,
iodine,
sulfur,
carbon, and
hydrogen.

Thus, the atom that is the most electronegative is fluorine (F).

Learn more about electronegativity here: brainly.com/question/24977425

#SPJ1

4 0

2 years ago

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

4 years ago

How far will an object move in 6 s if its average

Dafna11 [192]

Given:-

Time taken by the particle (t) = 6 s
Average speed (v) = 40 m/s

To Find: Distance (s) travelled by the particle.

We know,

s = vt

where,

s = Distance travelled,
v = Speed &
t = Time taken.

Putting the values,

s = (40 m/s)(6 s)

→ s = 240 m ...(Ans.)

6 0

3 years ago