All categories

3 years ago

Convert: 1) 2kg into gram 2) 5200m into km 3) 20cm into m

1 answer:

8 0

1) 2 kilograms converted into grams is 2000

2) 5200 meters converted into kilometers is 5.2

3) 20 centimeters converted into meters is 0.2

You might be interested in

An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac

elena-s [515]

First of all, we need to convert the angular speed from rev/min into rev/s:

$\omega_f=300 rev/min=5 rev/s$

The angular acceleration is the variation of angular speed divided by the time:

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{5 rev/s-0}{2 s}=2.5 rev/s^2$

And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

$\theta(t)=\frac{1}{2}\alpha t^2 =\frac{1}{2}(2.5 rev/s^2)(2 s)^2=5 rev$

so, 5 revolutions.

3 0

2 years ago

what has more momentum, a baseball traveling at 4 m/s or a baseball traveling at 16m/s ? and which has more energy ?

Anna007 [38]

We know the formulas for momentum and energy. But they both involve the mass of
the object, and we don't know the mass of the baseball. What can we do ?

It's not a catastrophe. The question only asks which one is bigger. If we're clever,
we can answer that without ever knowing how much the momentum or the energy
actually is. We know that both baseballs have the same mass, so let's just call it
' M ' and not worry about what it really is.

<u>Momentum of anything = (mass) x (speed)</u>
Momentum of the first baseball = (M) x (4 m/s) = 4M
Momentum of the second one = (M) x (16 m/s) = 16M
The second baseball has 4 times as much momentum as the first one has.

<u>Kinetic energy of anything = 1/2 (mass) x (speed squared)</u>
KE of the first baseball = 1/2 (M) x (4 squared) = 8M
KE of the second one = 1/2 (M) x (16 squared) = 128M
The second baseball has 16 times as much kinetic energy as the first one has.

3 0

3 years ago

An elastic rope is tied securely to a wall. An upward-displaced pulse is introduced into the rope. The pulse travels through the

Natasha2012 [34]

Answer:

Inverted (displaced downwards)

Explanation:

The pulse becomes INVERTED upon reflecting off the boundary with the wall. That is, an upward-displaced pulse will reflect off the end and return with a downward displacement. This inversion behavior will always be observed when the end of the medium is fixed, like this wall in this instance. This INVERSION BEHAVIOR can also be observed when the medium is connected to another more heavy or more dense medium. And in this case, when the pulse reaches the end of the medium, a portion of the pulse will reflect off the end and return with an inverted displacement. The heavier medium acts like a fixed end to cause the pulse to be inverted.

Summary: a pulse reaching the end of a medium becomes inverted whenever it either:

i. reflects off a fixed end,

ii. is moving in a less dense medium and reflects off a more dense medium.

3 0

3 years ago

Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from

Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

Density of the ice, ρ = 917 kg/m³
Initial pressure, P₁ = 1.0 bar
Final pressure, P₂ = 2.0 bar
Temperature, T = - 10 C
Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

$\Delta G = V(P_2-P_1) \\\\$

where;

V is the volume of the ice

$Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L$

Change in pressure;

$P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm$

The change in the Gibb's free energy per mole (G);

$\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G = 0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J$

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

3 0

2 years ago

Which of these is an example of projectile motion?

Usimov [2.4K]

Answer:

the answer is d I'm pretty sure

8 0

3 years ago