All categories

3 years ago

Does the inertia of an object change as the object’s velocity changes explain

1 answer:

8 0

Inertia is proportional to mass. It is a measure of the resistance to changes in velocity. Inertia is a property of mass and cannot change. Momentum changes as an object changes its velocity. Good luck on your assignment and have a great day! :D

You might be interested in

An instrument used to detect a static electric charge is called an

Eddi Din [679]

It is B. false that an instrument used to detect a static electric charge is called an ammeter. It is actually called an electroscope. Ammeter measures current.

7 0

3 years ago

Read 2 more answers

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

A box of mass 13kg sits on a table with a coefficient of static friction of 0.85. What is the maximum force of static friction?

Alexus [3.1K]

Answer:

108.3 N

Explanation:

The maximum force of static friction acting on the box is given by:

$F=\mu N$

where

$\mu = 0.85$ is the coefficient of static friction

N is the normal reaction of the table on the box

Since the box is in equilibrium along the vertical direction, the normal reaction N is equal to the weight of the box, so:

$N=mg=(13 kg)(9.8 m/s^2)=127.4 N$

And so, the maximum force of static friction is

$F=(0.85)(127.4 N)=108.3 N$

7 0

3 years ago

The angle of refraction rounded to the nearest whole number

Alika [10]

Answer:

$15^{\circ}$

Explanation:

Using Snell's law which is represented by

$n_1sin\theta_1 = n_2sin\theta_2$

Making $\theta_2$ the subject of the formula then

$\theta_2=sin^{-1}(\frac {n_1sin\theta_1}{n_2})$

Here $\theta_1$ and $\theta_2$ are the angles of incidence and refraction in water and air respectively

$n_1$ and $n_2$ are refraction index

Substituting 1.0003 for $n_1$ and 1.33 for $n_2$ then $20^{\circ}$ for $\theta 1$ we obtain

$\theta_2=sin^{-1}(\frac {1.0003\times sin 20^{\circ}}{1.33})=14.90606875^{\circ}\approx 15^{\circ}$

6 0

3 years ago

when a satellite is a distance d from the center of the earth, the force due to gravity on the satellite is F. what would be the

swat32

F₁ = c / d²
F₂ = c / (3d)²

F₁/F₂ = 3² = 9

F₂ = 1/9 F₁

5 0

3 years ago