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zvonat [6]
2 years ago
11

Engineers are trying to improve a race car. Their goal is to increase the acceleration of the car using the same engine. Which c

hange shows the best application of Newton’s laws of motion to achieve this goal?
increasing the inertia of the car
decreasing the reaction force
increasing the action force
decreasing the mass of the car
Physics
1 answer:
Andru [333]2 years ago
4 0

To increase the acceleration of the car using the same engine, the   mass of the car must be decreased.

<h3>What is Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that path unless acted upon by an external force.

The first law is also called the law of inertia because it depends on the mass of the object. The greater the mass, the greater the inertia and more reluctant the object will be to move.

Thus, to increase the acceleration of the car using the same engine, the   mass of the car must be decreased.

a = F/m

Learn more about Newton's law here: brainly.com/question/25545050

#SPJ1

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Explanation:

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3 years ago
A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio
LuckyWell [14K]

Answer:

U = √Rg/sin2θ

Explanation:

Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.

Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.

According to the range formula,

R = U²sin2θ/g

Cross multiplying we have;

Rg = U²sin2θ

Dividing both sides by sin2θ, we have;

U² = Rg/sin2θ

Taking the square root of both sides we have;

√U² = √Rg/sin2θ

U = √Rg/sin2θ

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3 0
3 years ago
Solve the inequality. x/3 is greater than or equal to - 6. a. x ≥ –9 b. x ≥ 9 c. x ≥ –18 d. x ≤ –18
NISA [10]
To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS. 

~ x greater than or equal to -18

(C)
6 0
3 years ago
View this and help out??
Zina [86]
I would say b as well. I’m sorry if it’s wrong
5 0
3 years ago
A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe
mezya [45]

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

6 0
3 years ago
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