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:
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:
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.
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Keywords: speed, distance, kinematics
The work done is
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement of the object
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, , and so the work done is
Learn more about work:
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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
= U = m g h
Initial mechanical energy, in the lower part of the track
Em₀ = K = ½ m v²
Em=
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
= M v1f + m v
p₀ =
M v₁₀ = M + m v
As the shock is elastic the kinetic energy is conserved
K₀ =
½ M v₁₀² = ½ M ² + ½ m v²
Let's write the system of equations
M v₁₀ = M + m v
M v1₁₀² = M ² + m v²
We cleared v1f in the first we replaced in the second
= (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
2 m v₁₀ - v (m + 1) m/ M = 0
v₁₀ = v (m +1) / (2M)
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
Explanation:
Mass of bumper cars,
Initial speed of car A,
Initial speed of car Z,
Final speed of car A after the collision,
We need to find the velocity of car Z after the collision. Let it is equal to . Using the conservation of momentum as :
So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.