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3 years ago

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car's acceleration?

Physics

1 answer:

ElenaW [278]3 years ago

7 0

Answer: $2m/s^{2}$

Explanation:

Assuming the car is moving only in one direction (to the South), its acceleration $a$ is given by the following formula, which expresses the variation of the velocity in time:

$a=\frac{\Delta V}{\Delta t}$ (1)

Where:

$\Delta V=V_{2}-V_{1}=40m/s-10m/s=30m/s$ is the variation in velocity

$\Delta t=t_{2}-t_{1}=15s-0s=15s$ is the variation in time

Rewritting (1) with the calculated values:

$a=\frac{30m/s}{15s}$ (2)

$a=2m/s^{2}$ This is the acceleration of the car

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a. Give an example of a galvanic cell. What kind of reaction occurs in a galvanic cell? b. If one electrode in a galvanic cell i

Burka [1]

Answer:

a) Batteries and fuel cells are examples of galvanic cell

b) Ag-cathode and Zn-anode

c) Cell notation: Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)

Explanation:

a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.

A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.

b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.

In the given case, the E⁰ values from the standard reduction potential table are:

E⁰(Zn/Zn2+) = -0.763 V

E°(Ag/Ag+)=+0.799 V

Therefore, Ag will be the cathode and Zn will be the anode

c) In the standard cell notation, the anode half cell is written on the left followed by the salt bridge '||' and finally the cathode half cell to the right.

Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)

4 0

3 years ago

¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?

coldgirl [10]

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

$s=0+\frac{1}{2}(5)(10)^2=250 m$

3 0

3 years ago

4. The chart shows melting and boiling points of

kondor19780726 [428]

Answer:

The correct option is A

Explanation:

Firstly, it should be noted that the freezing point of a substance can be assumed to be melting point of that substance because a substance will normally change from liquid to solid (freezes) at the same point it changes from solid to liquid (melts). For example, water freezes at 0°C and also starts melting at 0°C.

Thus, the substance with the lowest melting point among the substances mentioned in the question is alcohol (ethanol) with the melting point of -114°C. Hence, <u>ethanol also has the lowest freezing point thereby freezing at the lowest temperature.</u>

6 0

3 years ago

In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo

Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144 × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

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3 years ago

A tennis ball is hit into the air with a racket. When is the ball’s kinetic energy the greatest? Ignore air resistance.

Lady bird [3.3K]

Answer B is the correct answer

We know that kinetic energy $E = \frac{1}{2} mv^2$ , where m is the mass of object and v is the velocity of object.

In this case only velocity is the variable, mass remains constant.

So point having higher velocity has higher kinetic energy.

When it leaves the racket, the ball will be having a certain height, but just before it reaches the ground it will not having any height. So maximum velocity of ball is at that time when it reaches just above the ground.

So option B is the correct answer.

4 0

4 years ago