A car has a mass of 8000 kg and a kinetic energy of 24,000 J. What is its speed?

Through the process of blank

maksim [4K]

Answer:

Subduction, Trench, Mantle

Explanation:

8 0

2 years ago

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A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t

Sladkaya [172]

Answer:

The new pressure = 5.64 atm.

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

making P₂ the subject of the equation

P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters

Substituting these values into equation 2,

P₂ = (2.5×900×336)/(298×450)

P₂ = 756000/134100

P₂ = 5.64 atm.

Thus the new pressure = 5.64 atm.

5 0

2 years ago

Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end

wariber [46]

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

4 0

2 years ago

An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south

vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

vₐ = vₙ + vₐₙ

vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

8 0

2 years ago

Our solar system is located very close to the center of the Milky Way galaxy, outside of the spiraling arms.

inna [77]

FALSE
IF IT WAS THEN IT WOULd BE TRUE

4 0

3 years ago

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