What is the net magnetic flux through any closed surface?

How might burning fossil fuels affect the composition of gases in the atmosphere?

No "might". The amount of CO2 in the atmosphere HAS gone up since the start of industrialisation as the result of burning fossil fuels.

5 0

2 years ago

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago

A 15 kg object and a 18 kg object are connected by a massless compressed spring and

MatroZZZ [7]

Hi there!

This is an example of a recoil collision.

Using the conservation of momentum:

$p_i = p_f$

The initial momentum is 0 kgm/s (objects start from rest), so:

$p_f = 0$

We are given that the 15 kg block has a velocity of 12 m/s to the left, so:

$m_1v_1' + m_2v_2' = 0 \\\\15(-12) + 18v_2' = 0 \\\\$

Solve for v2':

$18v_2' = 180 \\\\v_2' = \boxed{10 m/s}$

5 0

2 years ago

In Young’s two slit experiment, the first dark fringe above the central bright fringe occurs at an angle of 0.44˚. What is the r

melomori [17]

Answer:

d / λ = 26.7

Explanation:

In Young's double slit experiment, constructive interference is described by the expression

d sin θ = m λ

In the case of destructive interference we must add half wavelength (λ/2)

d siyn θ = (m + ½) λ

Let's clear

d / λ = (m + ½) / sin θ

Let's calculate

d / λ = (2+ ½) / sin 5.4

d / λ = 5 / (2 sin 5.4)

d / λ = 26.7

8 0

2 years ago

A particle of mass m moves under a conservative force with potential energy V ( x )= cx/(x2+a2),where c and a are positive const

Anvisha [2.4K]

Answer:

The position of stable equilibrium is -a

And the period of small oscillations must be: c/(ma^3)

Explanation:

Since the potential is:

$V(x) = \frac{c x}{a^2+x^2}$

We first look for a position of stable equilibrium. This posiiton must satisfy two considtions, that the first derivative of the potential must vanish at this point and the second derivative must be positive.

$V'(x) = \frac{c}{a^2+x^2}-\frac{2 c x^2}{\left(a^2+x^2\right)^2}$

Which vanish for

x = a ; x =-a

The second derivative of V(x) is:

$V''(x) = \frac{8 c x^3}{\left(a^2+x^2\right)^3}-\frac{6 c x}{\left(a^2+x^2\right)^2}$

And:

$V''(a) = -\frac{c}{2 a^3}\\V''(-a) = \frac{c}{2 a^3}\\$

Therefore:

a)

The position of stable equilibrium is -a

And the period of small oscillations must be:

$\omega = \sqrt{2 V''(-a)/m} = \sqrt{\frac{c}{a^3 m}}$

(c/(ma^3))^1/2

b)

Let's find the maximum amplitude if the particle starts at this point with velocity v

If this is the case, the total energy will be:

(mv^2)/2

And the maximum amplitude will be

x = a^3/c mv^2 = (m v^2 a^3)/ c

7 0

3 years ago