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Furkat [3]
3 years ago
12

A car is traveling at 75 m/s. 50 seconds later it is traveling at 25 m/s. What is the car’s acceleration?

Physics
1 answer:
scoray [572]3 years ago
5 0

Answer:

a = -1 m/s^2

Explanation:

Vi = 75 m/s

Vf = 25 m/s

t = 50 s

Plug those values into the following equation:

Vf = Vi + at

25 = 75 + 50a

---> a = -1 m/s^2

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D all of the above (:
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How does the end point differ from the equivalence <br>point of a titration?​
Gwar [14]

<u>Answer:</u>

<em>Equivalence point and end point are terminologies in pH titrations and they are not the same. </em>

<u>Explanation:</u>

In a <em>titration the substance</em> added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the <em>acid and base titrated</em> determines the nature of the final solution.

At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation.  The nature of the final solution determines the <em>pH at equivalence point. </em>

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3 0
3 years ago
Chemical substances in food that have specific functions in your body are called?
andreev551 [17]

Answer:

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4 0
3 years ago
A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?
VLD [36.1K]
The initial kinetic energy of the car is
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Then, the velocity of the car is decreased by half: v_2 =  \frac{v_1}{2}
so, the new kinetic energy is
E_2 =  \frac{1}{2}mv_2 ^2 =  \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
E_2 =  \frac{1.9 \cdot 10^3 J}{4}=475 J
5 0
4 years ago
Suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg.
ladessa [460]

The other 4 kg of mass may have departed the scene
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3 years ago
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