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creativ13 [48]
2 years ago
10

Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is

Physics
1 answer:
s2008m [1.1K]2 years ago
6 0

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: \frac{∆v}{t}

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 \frac{feet}{second} then sped up to 5 \frac{feet}{second}, then the change in my speed would be 2 \frac{feet}{second}, because I started walking 2 \frac{feet}{second} faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 \frac{meters}{second} to 16 \frac{meters}{second}  in 4 seconds. What would the acceleration be? Let's set up an equation:

a = \frac{∆v}{t}

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 \frac{meters}{second} by 4 \frac{meters}{second}. That makes sense, right? Back to the equation.

a = \frac{∆v}{t}
a = \frac{16-4}{4}

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = \frac{12}{4}

a = 3 (\frac{meters}{second^{2}})

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit \frac{\frac{meters}{second}}{second}, which can be simplified down to \frac{meters}{second^{2} })

Let me know if you need clarification on anything I explained here!
- breezyツ

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What is the normal force and gravitational force on a 55 kg person standing stationary?
Alex777 [14]

Answer:

539.55 N for both

Explanation:

F=ma

Mass is given at 55kg

And when finding gravitational force gravity's constant is an acceleration of 9.81m/s

F=(55)(9.81)

F= 539.55 N (for both)

Normal force cancels out gravitational force that's why we don't fall throught the floor.

We push down on the floor due to gravitational force but normal force keeps us upright pushing an equal amount back.

4 0
3 years ago
A motorcyclist goes around an un-banked (i.e., flat) circular turn of radius 31m, at a constant speed of 110km/hr (convert this
shepuryov [24]

Answer:

\mu_{s} = 3.071

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

Explanation:

From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:

\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R} (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

N - Normal force, measured in newtons.

m - Mass of the motocyclist, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

v - Speed of the motorcyclist, measured in meters per second.

R - Radius of the circular turn, measured in meters.

The static coefficient of friction is cleared in (Eq. 1):

\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}

From (Eq. 2) we get that normai force is:

N = m\cdot g

And we expand the resulting expression in (Eq. 1):

\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}

\mu_{s} = \frac{v^{2}}{g\cdot R} (Eq. 3)

If we know that v = 30.556\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and R = 31\,m, the expected static coefficient of friction is:

\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}

\mu_{s} = 3.071

This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.

4 0
3 years ago
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