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solniwko [45]
3 years ago
13

If the car’s speed decreases at a constant rate from 60 mi/hmi/h to 50 mi/hmi/h in 3.0 ss, what is the magnitude of its accelera

tion, assuming that it continues to move in a straight line?
Physics
1 answer:
Kitty [74]3 years ago
7 0

Answer:

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

Explanation:

Hi there!

The acceleration is defined as the change in velocity in a time:

a = Δv / Δt

Where:

a = acceleration.

Δv = change in velocity  = final velocity - initial velocity.

Δt = elapsed time.

In this case:

Initial velocity = 60 mi/h

final velocity = 50 mi/h

elapsed time = 3.0 s

Let´s convert the time unit into h:

3.0 s · 1 h /3600 s = 1/1200 h

Now, let´s calculate the acceleration:

a = Δv / Δt

a = (50 mi/h - 60 mi/h) / 1/1200 h

a = -1.2 × 10⁴ mi/h²

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

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3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
Un prisma de cemento pesa 2500 N y ejerce una presión de 125 Pa, ¿cuál es el valor del área en la cual se apoya?
Eduardwww [97]

Answer:

Area = 20 m²

Explanation:

Given the following data;

Force = 2500 N

Pressure = 125 Pa

To find the area on which it rest;

Mathematically, pressure is given by the formula;

Pressure = \frac {Force}{area}

Making area the subject of formula, we have;

Area = \frac {Force}{pressure}

Substituting into the formula, we have;

Area = \frac {2500}{125}

Area = 20 m²

3 0
3 years ago
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