The energy stored per unit volume of the inductor is called the energy density of the magnetic field. Why?

A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the v

Rasek [7]

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

$[m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a$ Filling in:

$[1200(4.5)+2100(0)]=[(1200+2100)v]$ which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

6 0

2 years ago

A defibrillator passes 14.0 A of current through the torso of a person for 0.0300 s. How much charge moves in coulombs?

krek1111 [17]

<h2>Answer:</h2>

4.2 C

<h2>Explanation:</h2>

The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.

i.e

Q = I x t

Where;

I = current = 14.0A

t = time taken = 0.0300s

Substituting the values of I and t into the equation above gives

Q = 14.0 x 0.0300

Q = 4.2 C

Therefore quantity of charge moving is 4.2 C

4 0

3 years ago

an aircraft has a liftoff speed of 53 m/s. what is the minimum constant acceleration an airplane must have to reach that takeoff

xxMikexx [17]

Answer:

$a=3.34\ m/s^2$

Explanation:

<u>Accelerated Motion </u>

It refers to the motion of objects in which velocity is not constant over time. If the change of the velocity occurs at the same rate, then we say it's uniformly accelerated. Being $v_o$ = initial speed, $v_f$ = final speed, a= constant acceleration, x= distance traveled

Then, the scalar relation between them is

$v_f^2=v_o^2+2ax$

The aircraft needs to reach a liftoff speed of 53 m/s from rest (assumed) having only 420 meters to do so. We can compute the acceleration by solving for a

$\displaystyle a=\frac{v_f^2-v_0^2}{2x}$

$\displaystyle a=\frac{53^2-0^2}{2(420)}$

$\boxed{a=3.34\ m/s^2}$

6 0

3 years ago

If a negatively charged particle enters a region of uniform magnetic field which is perpendicular to the particle's velocity, wi

lys-0071 [83]

Answer:

Same

Explanation:

While moving through a magnetic field in a direction perpendicular to a B-field, a continuous force experienced by a charged particle. If this magnetic field remains uniform, the force exerted also remains same and hence the velocity with which the particle is moving remains same. However, the particle is forced to move on a curved path until it forms a complete circle.

Hence, the kinetic energy remains the same because the speed is same

3 0

2 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0

3 years ago