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myrzilka [38]
3 years ago
10

A car designer is designing a new model of a popular car. he wants to use the same engine as in the old model, but improve th ne

w car's acceleration. use newton's second law to explain how to improve the car's acceleration. use newton's second law to explain how to improve the cars acceleration without redesigning the engine.
Physics
1 answer:
Doss [256]3 years ago
5 0
It depends the amount if mass the object haves and the amount of force it is applied to the object.
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Which process allows a population Who become better adapted for survival in a specific environment
GalinKa [24]
Natural selection, it's simply when one animal outsmarts the rest and lives on to reproduce and carry on it's traits, when the animals who have not adapted well do not have the chance to reproduce, and the population dies off.
7 0
3 years ago
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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the
Paul [167]

Answer:

<em>a) 3344 N</em>

<em>b) 3344 N</em>

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = <em>3344 N</em>

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is <em>3344 N </em>

7 0
3 years ago
€£¥|~]£892ddddddddddd and I are going to go shopping with you and Toad and I can make it to the park and I will give you a call
storchak [24]

Answer:

what is this I don't get a thing

7 0
3 years ago
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The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular mo
emmainna [20.7K]

Answer:

b. 84 minutes

Explanation:

a_c=g = Centripetal acceleration = 9.81 m/s²

r = Radius of Earth = 6378.1 km

v = Velocity

Centripetal acceleration is given by

a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s

Time period is given by

T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes

The time period of Earth’s rotation would be 84.43835 minutes

7 0
3 years ago
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