A car is moving at 30.0 km/h when it accelerates at 2.0 m/s for 3.6 s. what is the car final speed?

Physics

1 answer:

Lana71 [14]4 years ago

3 0

Answer:

wrong question

Explanation:

unit of acceleration is written in m/s which is wrong

You might be interested in

Which would best be explained using Newton's Third Law?

Maksim231197 [3]

A)a gun moves backward while firing a bullet.

Explanation:

A gun moving backward while firing a bullet is a good example of newton's third law in action.

Newton's third law of motion states that "Action and reaction forces are equal and opposite in direction".

In shooting a gun, the bullet pulls forward by the action of the recoil down the barrel. This is the action force. The backward pull of the gun is the reactive force.

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

6 0

3 years ago

A gas is confined in a cylinder with a piston. What happens when work is done on the gas.

goblinko [34]

It combusts in to a vapor

5 0

3 years ago

Read 2 more answers

Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on

Soloha48 [4]

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

$B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\$

Next, you use the formula for the magnetic force produced by the wires:

$\vec{F_B}=I\vec{L}\ X \vec{B}$

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

$\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}$