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3 years ago

A small car and a large truck are both driving south at 40 km/h. Which of the following is true?

1 answer:

7 0

The right answer for the question that is being asked and shown above is that: "B.The small car has less momentum because its mass is less than the mass of the truck." A small car and a large truck are both driving south at 40 km/h. The statement that is true is that t<span>he small car has less momentum because its mass is less than the mass of the truck. </span>

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A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago

A 10-meter-long spear is thrown at a relativistic speed through a 10-meter-long pipe (both measured when at rest.) when the spea

Mekhanik [1.2K]

The correct answer is a

8 0

3 years ago

I select true helllpppp me

yanalaym [24]

Answer:

your right answer is true

hope it helps you

3 0

2 years ago

Read 2 more answers

Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum in

Nikitich [7]

Answer:22 m/s

Explanation:

Given

launch angle $\theta =60^{\circ}$

height of goal $h=4.55\ m$

and horizontal distance $x=40\ m$

Suppose initial speed is $u$

Trajectory of a Projectile is given by

$y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }$

substituting the values we get

$4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }$

$4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}$

$\frac{31,360}{u^2}=69.28-4.55$

$\frac{31,360}{64.73}=u^2$

$u^2=484.47$

$u=22.01\ m/s$

So, initial launch speed is $22\ m/s$

4 0

3 years ago

in case one, a car speeds up from zero m/s to 15 m/s. in case two, the same car speeds up from 15 m/s to 30 m/s. the mass of the

GrogVix [38]

Three time the energy was needed than in case one.

What is energy?

In physics, energy is the quantifiable property that is transmitted to a body or a physical system and is visible in the form of heat and light. The law of conservation of energy states that energy can be converted in form but cannot be created or destroyed.

Case 1:

Vo = 0 m/s

V1 = 15 m/s

Case 2:

Vo = 15 m/s

V1 = 30 m/s

Change in KE = 1/2 × m × (V1^2 - Vo^2)

KE1 = 1/2 × 1000 × (15^2 - 0)

= 112.5 kJ

KE2 = 1/2 × 1000 × (30^2 - 15^2)

= 1/2 × 1000 × 675

= 337.5kJ

Case 1 has an increase of 112.5 kJ energy while Case 2 has an increase of 337.5kJ.

To know more about energy click

brainly.com/question/582060

#SPJ4

7 0

1 year ago