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3 years ago

¿cual es la velocidad de un auto que recorre 4566 metros en 4 minutos? expresar en km\h

Physics

1 answer:

Svetlanka [38]3 years ago

3 0

Explanation:

(4566 m / 4 min) × (1 km / 1000 m) × (60 min / h) = 68.49 km/h

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Is it possible for an object to have a charge of 4.8x10-9 C? Why or why not?

Gelneren [198K]

I have absolutely no clue

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3 years ago

At a certain point in space, there is a potential of 800 V relative to zero. What is the potential energy of the system when a +

sammy [17]

To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

$V = \frac{k_e q}{r}$

$k_e$ = Coulomb's constant

q = Charge

r = Radius

At the same time

$U = \frac{k_e q_1q_2}{r}$

The values of variables are the same, then if we replace in a single equation we have this expression,

$U = Vq$

If we replace the values, we have finally that the charge is,

$V = 800V$

$q = 1\mu C$

$U = (800V)(1*10^{-6}C)$

$U = 8*10^{-4}J$

Therefore the potential energy of the system is $8*10^{-4} J$

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3 years ago

Which are examples of short-term environmental change? Check all that apply.

d1i1m1o1n [39]

Answer:

tsunamis,El Nino,and volcanic eruptions

Explanation:

all short term.

8 0

3 years ago

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The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

3 years ago

Mars, which has a radius of 3.4 × 106 m and a mass of 6.4 × 1023 kg, orbits the Sun, which has a mass of 2.0 × 1030 kg at a dist

Elena L [17]

Answer:

Tangential speed of Mass revolution is greater

Explanation:

In the above question, we are given the following values:

Radius of Mars = 3.4 × 10^6 m

Mass of Mars = 6.4 × 10^23 kg

Mass of the sun = 2.0 × 10^30 kg at a distance of 2.3 × 10^11m.

Mar’s revolution around the sun = 687 days.

Time of Revolution = 24 hours and 37 minutes to complete one revolution.

Step 1

Find the tangential speed of rotation of Mars

Rotation of Mars = 2 × π × r

Radius of Mars = 3.4 × 10^6 m

Rotation of Mars = 2 × π × 3.4 × 10^6 m

Rotation of Mars = 21362830.044m

= 21.4 × 10^6 m

Tangential Speed of rotation for Mars = Rotation of Mars/ time

Time = 24 hrs 37 minutes

Converting to seconds = 88620s

Tangential Speed of rotation = 21.4 × 10^6 m/ 88620s

= 241.48047845m/s

= 241.5 m/s

Step 2

revolution of Mars = 2 × π × d

d = 2.3 × 10^11m.

= 2 × π × 2.3 × 10^11m = 1.45 × 10^12m

Revolution of Mars = 687 days

Converting 687 days to seconds = 59356800s

Tangential Speed of Revolution of Mars = Revolution of Mars/ times

Tangential Speed of Revolution = 1.45 × 10^12m/595356800s

Tangential Speed of Revolution = 24420m/s

Comparing the Tangential Speed of Rotation with the Tangential speed of Revolution, we can see the GREATER ONE IS THE TANGENTIAL SPEED OF REVOLUTION.

8 0

3 years ago

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