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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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1ST ANSWER WILL BE BRAINLIEST
zloy xaker [14]
All of the above. They all right
4 0
4 years ago
Read 2 more answers
A 36.3 kg cart has a velocity of 3 m/s. How much kinetic energy does the object have?
uysha [10]

Answer:

163.35

__________________________________________________________

<u>We are given:</u>

Mass of the object (m) = 36.3 kg

Velocity of the object (v) = 3 m/s

<u>Kinetic Energy of the object:</u>

We know that:

Kinetic Energy = 1/2(mv²)

KE = 1/2(36.3)(3)²            [replacing the variables with the given values]

KE = 18.15 * 9

KE = 163.35 Joules

Hence, the cart has a Kinetic Energy of 163.35 Joules

7 0
3 years ago
Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
4 years ago
5. How does a jack make changing a tire easier?
strojnjashka [21]

Answer: An jack makes changing a tire easier because it lifts up the car to get the tire off of the ground.

Explanation:

4 0
3 years ago
Magnet A has twice the magnetic field strength of magnet B and pulls on magnet B with a force of 100 N. The amount of force that
son4ous [18]

The force exerted by the magnetic in terms of the magnetic field is,

F\propto B

Where B is the magnetic fied strength and F is the force.

Thus, if the magnetic A has twice magnetic field strength than the magnet B,

Then,

B_A=2B_B

Thus, the force exerted by the magnet B is,

\begin{gathered} F_B\propto B_B \\ F_B\propto\frac{B_A}{2} \\ F_B=\frac{F_A}{2} \\ F_B=\frac{100}{2} \\ F_B=50\text{ N} \end{gathered}

Thus, the force exerted by the magnet B on magnet A is 50 N.

The force exerted by the magnet A exerts on the magnet B is exactly 100 N as given.

Hence, the option B is the correct answer.

3 0
1 year ago
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