Ask question

Ask question

All categories

3 years ago

11

Physical science question please help

1 answer:

emmainna [20.7K]3 years ago

8 0

A compound contains atoms of different elements chemically combined together in a fixed ratio. An element is a pure chemical substance made of same type of atom. ... Compounds contain different elements in a fixed ratio arranged in a defined manner through chemical bonds.

You might be interested in

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

3 years ago

Read 2 more answers

Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a

Levart [38]

He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements

6 0

3 years ago

Read 2 more answers

What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?

ratelena [41]

= 80 N/40 kg
= 2 m/s 2

4 0

3 years ago

Which of these is NOT a possible type of energy transformation?

Orlov [11]

to be franc i really think the answer is B

6 0

3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

3 years ago