Answer:
The car manufacturers could increase bore of the cylinders, place the engine in the center or back of the car, add 1 to 2 turbochargers, and lower the center of gravity of the vehicle to increase traction.
Explanation:
Turbochargers would be recommended because they significantly increase both the torque of the engine as well as the amount of horses powering the car while also increasing original efficiency both with and without the additional power. Weight adjustment allows for lightweight vehicles with good traction. This is important to both keep control of the car under acceleration, but it also makes the vehicle more efficient due to the now sheddable unnecessary weight. A more obvious approach would be to increase the base horsepower and torque of the engine by increasing the bore of the cylinders and the weight of the pistons. This acts as an inertial lever, because the extra piston weight will drag the crankshaft faster. This could also be achieved by taking away piston weight, but this could be catastrophic should a piston slip.
Answer:
An interaction of one object with another object results in a force between the two objects. Thus, at-least two objects must interact for a force to come into play.
No, since their gravity is powerful enough to keep them together even while the universe expands as a whole. Space is not expanding within clusters of galaxies.
<h3>What is a galaxy?</h3>
A galaxy is a massive clump of gas, dust, and billions of stars and their solar systems bound together by gravity.
No, since their gravity is powerful enough to keep them together even while the universe expands as an entire.
Hence,space is not expanding within clusters of galaxies.
To learn more about the galaxy, refer to the link;
brainly.com/question/2905713
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Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
