To increase a car's acceleration, the engineers should consider decreasing the car's mass and/or increasing the force that the engine provides. Therefore, options A and D are the correct answers.
Acceleration is directly proportional to the net force acting on an object and inversely proportional to its mass. This means that a smaller mass or a larger force will result in a greater acceleration. Therefore, reducing the mass of the car will require less force to achieve the same acceleration, or the same force will result in a greater acceleration.
To decrease the mass of the car, the engineers could consider using lighter materials for the car's body and frame or removing unnecessary components. On the other hand, to increase the force that the engine provides, the engineers could consider upgrading the engine or increasing the fuel intake.
Option B, which suggests decreasing the force that the engine provides, would have the opposite effect and result in a slower acceleration. Option C, which suggests increasing the mass of the car, would also have the opposite effect and require more force to achieve the same acceleration.
Option E, which suggests increasing the top velocity the car can travel, is not directly related to acceleration. Top velocity refers to the maximum speed that a car can reach, whereas acceleration refers to the rate of change of velocity. While increasing the car's top velocity may indirectly affect acceleration, it is not a direct solution to increasing acceleration.
Option A and D.
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a bus with a maximum speed of 20m/s takes 21sec to travel 270m from stop to stop. Its acceleration is twice as great as its deceleration. Find: A, the acceleration B, the distance traveled at maximum speed
Answer:
We can use the kinematic equations to solve this problem.
Let a be the acceleration and d be the distance traveled at maximum speed.
First, we can use the equation:
d = vt + (1/2)at^2
where v is the maximum speed, t is the time traveled at maximum speed, and a is the acceleration.
We know that the bus takes 21 seconds to travel 270 meters, so the time traveled at maximum speed is:
21 - 2a = t
We also know that the acceleration is twice as great as the deceleration, so we can write:
a = 2d
Then, we can substitute these expressions into the first equation:
d = (20)(21 - 2a) + (1/2)(2d)(21 - 2a)^2
Simplifying and solving for d, we get:
d = 210 - 5.25a + 0.25a^2
To find the acceleration, we can use the fact that the maximum speed is reached at the midpoint of the trip, so the distance traveled at maximum speed is half the total distance:
d = 1/2(270)
Solving for d, we get:
d = 135
Substituting this value into the equation for d, we get:
135 = 210 - 5.25a + 0.25a^2
Simplifying and solving for a, we get:
a = 8 m/s^2
Finally, we can use the equation:
d = vt
to find the distance traveled at maximum speed:
d = (20)(21 - 2a)
Substituting the value of a, we get:
d ≈ 188.4 m
Therefore, the acceleration of the bus is 8 m/s^2 and the distance traveled at maximum speed is approximately 188.4 meters.
What is the average acceleration of an
automobile moving from 0 to 60 m/s in 3.0
seconds?
a. 90 m/s²
b. -20 m/s²
c. -90 m/s²
d. 20 m/s²
Answer:
D
Explanation:
Using the formula for average acceleration, which is:
average acceleration = (final velocity - initial velocity) / time
Where the initial velocity is 0 m/s and the final velocity is 60 m/s, and the time is 3 seconds, we get:
average acceleration = (60 m/s - 0 m/s) / 3 s = 20 m/s²
Therefore, the answer is (d) 20 m/s².
a cello plays a c# note with a frequency of 346.2hz. if the speed of sound in air is 225m/s, what is the wavelength of this note?
The wavelength of the note produced by the cello is 0.65 m.
The frequency of the note played by cello, f = 346.2 Hz
Speed of the sound in air, v = 225 m/s
Speed of a wave is calculated by taking the product of its frequency and wavelength.
The expression for the speed of the sound wave is given by,
v = fλ
where λ is the wavelength of the note produced.
Therefore, the wavelength of the note produced by the cello,
λ = v/f
Applying the values of v and f,
λ = 225/346.2
λ = 0.65 m
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Two uniform solid spheres, each with mass 0.852 kg
and radius 8.00×10−2 m are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant 153 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring.
Assume that the motion of the center of mass of the spheres is simple harmonic. Calculate its period.
The period of the simple harmonic motion of the center of mass of the two spheres is approximately 0.770 seconds.
To find the period of the simple harmonic motion of the center of mass of the two spheres, we need to use the equation for the period of a mass-spring system:
T = 2π√(m/k)
where T is the period, m is the total mass of the system (two spheres), and k is the spring constant.
First, we need to find the total mass of the system:
m = 2m1 = 2(0.852 kg) = 1.704 kg
where m1 is the mass of one sphere.
Next, we need to find the spring constant:
k = 153 N/m
Now, we can calculate the period:
2π√(1.704 kg/153 N/m) ≈ 0.770 s
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Pressure is applied from a gas cylinder to one side of a U-tube manometer with constant internal diameter “D", as shown below. The manometer is filled with fresh water (density = 999 kg/m3) and upon application of the pressure, the water column on one side rises by 4.00 cm relative to the other. Draw and label a free-body diagram representing the forces in the manometer. Calculate the pressure above atmospheric pressure applied by the gas.
Answer:
P = 39200.76
Explanation:
formula is:
P = pgh
P = pressure
p = density of fluid
g = acceleration due to gravity ( 9.81 m/s2 for earth)
h = height above the point we are looking for
P = pgh
P = 999x9.81x4
P = 39200.76
