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2 years ago

6

Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved

1 answer:

lesantik [10]2 years ago

7 0

Momentum is conserved when carts are collided on a slanting plane.

To find the answer, we need to know about the conversation of momentum.

<h3>What's the conversation of momentum?</h3>

Conservation of linear momentum says the total momentum before the collision and after the collision remains the same.
Mathematically, m1u1+m2u2 = m1v1+m2v2

<h3>How is the momentum conserved when collision occurs on a slanting plane?</h3>

On a slanting plane, the velocity has two components,

horizontal component
horizontal component Vertical component

So, its momentum has also similar two components.
The momentum is conserved along horizontal direction and vertical direction separately.

Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.

Learn more about the conversation of momentum here:

brainly.com/question/7538238

#SPJ4

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A baseball leaves a bat and flies upward and toward center field. Do not ignore air resistance. What forces are exerted on the b

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4 years ago

Which of the following is true concerning atoms?Atoms are made of even smaller substances called subatomic particles

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4 years ago

A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic

Margarita [4]

Answer:

$\lambda = 424.4 nm$

Explanation:

As we know by the formula of diffraction

$a sin\theta = N\lambda$

so we have

a = slit size

$\theta$ = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

$a sin40 = 1(630 \times 10^{-9})$

$a = 9.8 \times 10^{-7} m$

Now for another wavelength second minimum is at 60 degree angle

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3 years ago

Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

3 years ago