A 1500 kg car accelerates at 3 m/s^2. What is the net force acting on the car?

Answers

Answer 1

Answer:

4500N

Explanation:

Force =mass into acceleration

So 1500 into 3 you will get 4500


Related Questions

The Law of Conservation of Mass states that mass can be destroyed during a chemical change.

True
False

Answers

The law of conservation of mass states that mass can be destroyed during a chemical change is TRUE

all of the following describe the motion of an object except
mass
speed
velocity
acceleration ​

Answers

all describe an objects motion except : mass

Solar energy heats the surface of the Earth, including the ground, rocks, and even roadways. As the temperature of these surfaces increases, heat energy is released back into the atmosphere, forming a pocket of warm air. As the warm air pocket rises, it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere and is known as

Answers

Answer:

convection

Explanation:

I hope this helps

Which of the following is an example of an electromagnetic wave ?

Answers

Answer:

Its a wave

Explanation:

THis is one

Answer: a light wave

Explanation: a pex

Compare the inner planets with the outer planets.

Answers

Answer:

Inner Planets:

- Rocky

- Small

- Only 3 moons total

- Very close to each other (for space)

- Underweight

Outer Planets:

- Gassy oop-

- GIANT

- Over 200 moons total

- VERY far from each other

- Obese

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Good luck foe.

Lb.

I need help asap

What is a homopolar motor?
a. The first electric motor ever made.
b. A motor made by in the North Pole.
c. A motor used in homes in the winter.

Answers

Answer: The answer is A

Explanation:

Explain how a current is induced in the wire?

Answers

Answer:

If a coil of wire is placed in a changing magnetic field, a current will be induced in the wire. This current flows because something is producing an electric field that forces the charges around the wire. (It cannot be the magnetic force since the charges are not initially moving). ... that determines the induced current.

Three charges lie along the x-axis. One positive charge, q1 = 4.80*10^-18 C, is at x = 3.72 m, and another positive charge, q2 = 1.60*10^-19 C, is at the origin.
At what point on the x-axis must a negative charge, q3, be placed so that the resultant force on it is zero?

Answers

Answer:

The third charge needs to be placed at [tex]x \approx 0.57\; \rm m[/tex].

Explanation:

Both [tex]q_1[/tex] and [tex]q_2[/tex] would attract [tex]q_3[/tex].

These two electrostatic attractions need to balance one another. Hence, they need to be opposite to one another. Therefore, [tex]q_1[/tex] and [tex]q_2[/tex] need to be on opposite sides of [tex]q_3[/tex]. That is possible only if [tex]q_3 \![/tex] is on the line segment between [tex]q_1 \![/tex] and [tex]q_2 \![/tex].

Assume that [tex]q_3[/tex] is at [tex]x\; \rm m[/tex], where [tex]0 < x < 3.72[/tex] (in other words, [tex]q_3 \![/tex] is on the line segment between [tex]q_1[/tex] and [tex]q_2[/tex], and is [tex]x\; \rm m \![/tex] away from [tex]q_2 \![/tex].)

Let [tex]k[/tex] denote Coulomb's constant.

The magnitude of the electrostatic attraction between [tex]q_1[/tex] and [tex]q_3[/tex] would be:

[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}}[/tex].

Similarly, the magnitude of the electrostatic attraction between [tex]q_2[/tex] and [tex]q_3[/tex] would be:

[tex]\displaystyle \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].

The magnitudes of these two electrostatic attractions need to be equal to one another for the resultant electrostatic force on [tex]q_3[/tex] to be [tex]0[/tex]. Equate these two expressions and solve for [tex]x[/tex]:

[tex]\displaystyle \frac{k\cdot q_1 \cdot q_3}{(3.72 - x)^{2}} = \frac{k\cdot q_2 \cdot q_3}{x^{2}}[/tex].

[tex]\displaystyle \frac{q_1}{(3.72 - x)^{2}} = \frac{q_2}{x^{2}}[/tex].

[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1}[/tex].

[tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{q_2}{q_1} = \frac{1}{30}[/tex].

By the assumption that [tex](0 < x < 3.72)[/tex], it should be true that [tex](x > 0)[/tex] and [tex](3.72 - x > 0)[/tex]. Therefore, [tex]\displaystyle \frac{x}{(3.72 - x)} > 0[/tex].

Take the square root of both sides of the equation [tex]\displaystyle \frac{x^2}{(3.72 - x)^{2}} = \frac{1}{30}[/tex].

[tex]\displaystyle \sqrt{\frac{x^2}{(3.72 - x)^{2}}} = \sqrt{\frac{1}{30}}[/tex].

[tex]\displaystyle \frac{x}{3.72 - x} = \frac{1}{\sqrt{30}}[/tex].

[tex]\sqrt{30}\, x = 3.72 - x[/tex].

Therefore:

[tex]\left(1 + \sqrt{30}\right)\, x = 3.72[/tex].

[tex]\displaystyle x = \frac{3.72}{1 + \sqrt{30}} \approx 0.57[/tex].

Hence, [tex]q_3[/tex] should be placed at [tex]x \approx 0.57\; \rm m[/tex].

how does displacement, acceleration, time, and velocity affect motion?

Answers

Answer:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.

~Hoped this helped~

~Brainiliest?~

Answer:

We describe motion in terms of velocity and acceleration. Velocity: The rate of change of displacement of an object (displacement over elapsed time) is velocity. Velocity is a vector since it has both magnitude (called speed) and direction. ... Acceleration: The rate of change of velocity is acceleration.

Explanation:

Displacement is a vector which points from the initial position of an object to its final position. ... Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity.

Displacement is the vector difference between the ending and starting positions of an object. Velocity is the rate at which displacement changes with time. ... The average velocity over some interval is the total displacement during that interval, divided by the time.

Hope this helps      :)

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