Ask question

Ask question

All categories

3 years ago

15

Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m

ake the comparison?

1 answer:

tatyana61 [14]3 years ago

7 0

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

You might be interested in

On May 20, 1999, 37-year old Robbie Knievel, son of the famed daredevil Evel

Ostrovityanka [42]

R/RocketLeague - Every Painted Uncommon Wheels (Pics)
It's alive, IT'S ALIVE!

Tired of trading some of the new wheels and not having a pic of them? Want to collect some sweet wheels but you don't know which? Say no more! The complete collection of the painted 17 different uncommon wheels it's here.

Tables with images for every wheel are right above. Also, you can check the Excel worksheet version right here.

The wheels contained here are all the u

6 0

3 years ago

Why evaporation are called surface phenomenon?

Eva8 [605]

The water formed on the surface of the water evaporation loss (evaporation), consisting of plant transpiration water loss (transpiration) is called. Soil near the plant and the resulting water loss is called by evapotranspiration.

6 0

4 years ago

Read 2 more answers

Steve tunes an old-fashioned radio that has an antenna made from a 3.0-cm-long solenoid with a cross-sectional area of 0.50 cm2

sukhopar [10]

Answer:

$L = 1.88\times10^-^4 H$

Explanation:

It is given that,

Length of the coil, l = 3 cm = 0.03 m

Area of cross section of the coil, $A=0.5\ cm^2=0.00005\ m^2$

Number of turns in the coil, N = 300

inductance of the coil is

$L=\dfrac{\mu_oN^2A}{l}$

$L=\dfrac{4\pi\times 10^{-7}\times (300)^2\times 5\times 10^{-5}}{0.03}\\L = 1.88\times10^-^4 H$

So, the inductance of the coil is 1.88× 10⁻⁴H

5 0

3 years ago

The gravitational force

vivado [14]

<span>b. weakens as 1/d, where d is the distance between objects.</span>

3 0

3 years ago

Read 2 more answers

In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second pist

marin [14]

Answer:

847.45 N

Explanation:

F₁=force of exerted by smaller piston

F₂=force of exerted by larger piston=1.44×10⁴ N

A₁=Area of smaller piston= 2.62 cm =0.0265 m

A₂=Area of larger piston= 10.8 cm =0.108 m

Pressure exerted by both the pistons will be equal

$P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times 0.108^2}\pi\times 0.0262^2\\\Rightarrow F_1=847.45\ N$

Hence, force exerted to lift a 14400 N car is 847.45 N

8 0

4 years ago