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Readme [11.4K]
3 years ago
7

Alex rides her bike at 12.0 km/hr for 2.25 hr. what distance has alex traveled?

Physics
1 answer:
Ymorist [56]3 years ago
6 0

Answer: Alex traveled a distance of 27.0 km in 2.25 hours.  

Further Explanation:

Speed is how fast an object moves or how far an object travels per unit time. If the distance traveled and the total time of travel are known, the speed can be calculated using the formula:

speed \ = \frac{distance}{time}

In the problem, we are given:

speed = 12.0 km/hr

time = 2.25 hr

We are looking for the distance traveled by Alex which can be represented by the variable d.

We can solve for d by manipulating the speed formula to get the equation:

distance, d \ = \ (speed)(time)[/tex]

Plugging in our values for speed and time, we get the equation:

distance \ = \ (12.0 \ \frac{km}{hr})(2.25 \ hr)\\ \boxed {distance,d\ = \ 27.0 \ km}

Since the number of significant figures of the given is 3, the answer must be expressed with 3 significant figures, too.

Thus, the distance Alex traveled for 2.25 hours at a speed of 12.0 km/hr is 27.0 km.

Learn More

  1. Learn more about Velocity brainly.com/question/862972
  2. Learn more about Acceleration brainly.com/question/4134594
  3. Learn more about Distance - Time Graphs brainly.com/question/1378025

Keywords: speed, distance, kinematics

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zhannawk [14.2K]

The work done is 2.35\cdot 10^5 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, we have

F = 1080 N is the force applied on the car

d = 218 m is the displacement of the car

And assuming the force is applied parallel to the motion of the car, \theta=0^{\circ}, and so the work done is

W=(1080)(218)(cos 0^{\circ})=2.35\cdot 10^5 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
2 years ago
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

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Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

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We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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