Two students have the same velocity during a race. Colin has a mass of 80 kg while Kara has a mass of 80 kg. If Kara doubled her speed how does her new momentum compare to Colin’s?

Answers

Answer 1

Answer: Kara has twice the momentum as Colin

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Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009

The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
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The focal length of a concave mirror is 17.5 cm. An object is located 38.5 cm in front of this mirror. How far in front of the mirror is the image located?

Answers

Answer:

Explanation:

object distance u = 38.5 cm ( negative )

focal length f = 17.5 cm ( negative )

mirror formula

1 / v + 1 / u = 1 / f

1 / v - 1 / 38.5 = - 1 / 17.5

1 / v = - 1 / 17.5 + 1 / 38.5

= - 0 .03116

v = - 1 / .03116 = - 32 cm

Image will be formed in front of the mirror at 32 cm distance .

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm What is the distance between the two second-order minima?

Answers

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=(y_1)/(2)\n\Rightarrow \beta_1=(1.4)/(2)\n\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=(m_1\lambda D)/(d)\n\Rightarrow d=(m_1\lambda D)/(\beta_1)$

For second order

$\beta_2=(m_2\lambda D)/(d)\n\Rightarrow \beta_2=(m_2\lambda D)/((m_1\lambda D)/(\beta_1))\n\Rightarrow \beta_2=(m_2)/(m_1)\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\n\Rightarrow y_2=2(m_2)/(m_1)\beta_1\n\Rightarrow y_2=2(2)/(1)* 0.7\n\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answers

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=(GMm)/(r^2)$

Where $G=6.67*10^(-11)Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=(GMm)/(r^2)$

Which means:

$a=(GM)/(r^2)$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=(2GMd)/(r^2)$

$v=\sqrt{(2GMd)/(r^2)}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5*10^3m$ . With our values then we have:

$v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s$

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?

Answers

The acceleration of the particle is 38.87 kg.

Net magnetic field

The net magnetic field is calculated as follows;

$B_(net) = √(B_x^2 + B_y^2 + B_z^2) \n\nB_(net) = √(2^2 + 3^2 + 4^2) = 5.385 \ mT$

Magnetic force on the charge

The magnetic force on the charge is calculated as follows;

$F = qvB * sin(\theta)\n\nF = 5* 10^(-9) * 5* 10^3 * 5.385 * 10^(-3) * sin(120)\n\nF = 1.166 * 10^(-7) \ N$

Acceleration of the particle

The acceleration of the particle is calculated as follows;

$a = (F)/(m) \n\na = (1.166 * 10^(-7))/(3 * 10^(-9)) \n\na = 38.87 \ kg$

Learn more about magnetic force here: brainly.com/question/13277365

Explanation:

It is given that,

Charge on the particle, $q=5\ nC=5* 10^(-9)\ C$

Mass of the particle, $m=3\ \mu g=3* 10^(-6)\ g=3* 10^(-9)\ kg$

Magnetic field component, $B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT$

Net magnetic field, $B=√(2^2+3^2+4^2)=5.38\ mT=0.00538\ T$

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, $\theta=120$

Magnetic force is given by :

$F=qvB\ sin\theta$

$F=5* 10^(-9)* 5000\ m/s* 0.00538* sin(120)$

$F=1.16* 10^(-7)\ N$

Acceleration of the particle is given by, $a=(F)/(m)$

$a=(1.16* 10^(-7)\ N)/(3* 10^(-9)\ kg)$

$a=38.6\ m/s^2$

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.

Two identical metal spheres a and b are in contact. both are initially neutral. 1.0×1012 electrons are added to sphere a, then the two spheres are separated. you may want to review ( pages 639 - 641) . part a afterward, what is the charge of sphere a?

Answers

The charge on the sphere A and sphere B after they are separated is $\boxed{ - 80\,{\text{nC}}}$ each.

Further Explanation:

Given:

The number of electrons transferred to sphere $A$ is $1.0 * {10^(12)}$ .

Concept:

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $- 1.6 * {10^( - 19)}\,{\text{C}}$ .

So, the amount of charge carried by the $1.0 * {10^(12)}$ electrons is given as.

$\begin{aligned}Q&= \left( {1.0 * {{10}^(12)}} \right)\left( { - 1.6 * {{10}^( - 19)}} \right)\n&= - 1.6 * {10^( - 7)}\,{\text{C}}\n\end{aligned}$

Since the charge is disturbed equally on the two sphere, so the amount of charge carried by each sphere s half of the total charge.

$\begin{aligned}{Q_A}&= \frac{{ - 1.6 * {{10}^( - 7)}}}{2}\n&= - 8 * {10^( - 8)}\,{\text{C}}\n &= - 8{\text{0}}\,{\text{nC}}\n\end{aligned}$

Thus, the amount of the charge carried by each sphere after separating from each other is $\boxed{ - 80\,{\text{nC}}}$ .

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords: Metal spheres, two identical, in contact, neutral, charged, electrons, charge on electron, charge on metallic sphere, charge of sphere A.

The charge on the sphere A and sphere B after they are separated is $-80\mu C$ each

What is Charge?

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

The amount of charge carried by the electrons when reaches the spheres kept in contact with each other is first distributed equally on each sphere. Later as the spheres are moved away from one another, the charge on each sphere remains the same as it was when they were in contact.

The amount of charge on one electron is $-1.6* 10^(-19) \ C$