All categories

3 years ago

What is the unit and tool to measure the volume of a plastic block?

1 answer:

8 0

Unit is m^3 or metres cubed. You need to multiply the three dimensions of the block to get the volume.
A ruler can be used to measure the edges.

You might be interested in

Describe the principle of conversation of energy:

exis [7]

Answer: Energy can neither be created nor destroyed, rather it is converted from one form to another

Explanation:

The principle of conversation of energy explains how energy is conserved in nature by being converted from one form to another such that no energy is created nor destroyed.

Practical examples include:

- electrical pressing iron that converts electrical energy to heat energy

- solar panels that converts solar energy to electrical energy

- Car batteries that converts chemical energy to light energy etc

7 0

3 years ago

Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

6 0

3 years ago

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

3 0

2 years ago

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

2 years ago

Tcecarenko [31]

Explanation:

Melting point

Boiling point

Does the compound

dissolve in water?

Can the compound

conduct electricity?

8 0

2 years ago