All categories

3 years ago

Type one to two paragraphs describing the changes in potential and kinetic energy of the cart.

Chemistry

2 answers:

Deffense [45]3 years ago

8 0

Answer:

potential energy is energy that is stored in an object. Can also transfer into others forms of energy. Kinetic energy is energy an object has become of its motion.

Umnica [9.8K]3 years ago

3 0

Answer:

When the cart started at the top of the hill it was full Potential Energy. As the cart started to go down the hill it was gaining speed turning it into (PE/KE) Kinetic Energy. That transition occurred because Kinetic Energy only happens when something gains speed. As the cart hit the valley. (on poster) It was still full of Kinetic Energy. As the cart started moving up the last hill it was transitioning from Kinetic Energy, to Potential Energy, (KE, PE). Which then led to the cart coming to a stop on the second hill as Potential Energy. (PE)

Please mark me brainliest.

Hope I helped.

You might be interested in

What is the empirical formula for a compound that is 29.44\% calcium, 23.55% sulfur, and 47.01% oxygen? This compound is a commo

PSYCHO15rus [73]

Answer:

Empirical formula is CaSO₄.

Explanation:

Given data:

Percentage of calcium =29.44%

Percentage of sulfur = 23.55%

Percentage of oxygen = 47.01%

Empirical formula = ?

Solution:

Number of gram atoms of Ca = 29.44 / 40 = 0.74

Number of gram atoms of S = 23.55 / 32 = 0.74

Number of gram atoms of O = 47.01 / 16 = 3

Atomic ratio:

Ca : S : O

0.74/0.74 : 0.74/0.74 : 3/0.74

1 : 1 : 4

Ca : S : O = 1 : 1 : 4

Empirical formula is CaSO₄.

3 0

2 years ago

⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

4 0

3 years ago

The worst feeling is when you are hurting so badly inside but not a single tear comes out.

guapka [62]

Answer:

chemistry wow

Explanation:

8 0

2 years ago

An experiment was conducted and students observed that a solution had a hydroxide concentration of 7.0 x 10-2 .What is the pH? i

denis23 [38]

Answer:

The pH of the solution is basic

Explanation:

as one of the characteristics of the base is to increase the OH concentration in a solution so if the solution has low concentration of OH this means that the solution is basic

3 0

3 years ago

How to find change in enthalpy

arlik [135]

∆H = m x s x ∆T, where m is the mass of the reactants, s is the specific heat of the product, and ∆T is the change in temperature from the reaction.

5 0

2 years ago