An object moving at a constant velocity accelerates at a constant rate of

Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

$v=v_o+at$ (1)

$x=v_ot+\frac{1}{2}at^2$ (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

$v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s$

Next, you solve the equation (1) for the acceleration a:

$a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}$

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

$x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm$

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0

3 years ago

A technological device that can be used to see sound waves is an

Bond [772]

The answer is oscilloscope

4 0

3 years ago

show answer Incorrect Answer 33% Part (b) Find the radius of curvature, in meters, of the path of a proton accelerated through t

timofeeve [1]

The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

(c) ratio = 1800

Explanation: (a) <u>Potential</u> <u>difference</u> is defined as the energy a charged particle has between two points in a circuit. It is calculated as

$\Delta V=\frac{pe}{q}$

where

pe is potential energy

q is charge

and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

so

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

Potential difference of an electron to have speed of 6.2x10⁷m/s is $-109.44$ x 10²V.

(b) A particle has a circular motion when there is a magnetic force acting on it.

Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

$F_{mag} = F_{c}$

$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

5 0

4 years ago

A cannon, elevated at 40∘ is fired at a wall 300 m away on level ground, as shown in the figure below. The initial speed of the

Gre4nikov [31]

Vo = 89 m/s
angle: 40°

=> Vox = Vo * cos 40° = 89 * cos 40°

=> Voy = Vo. sin 40° = 89 * sin 40°

x-movement: uniform => x =Vox * t = 89*cos(40)*t

x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s

y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2

y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.

7 0

3 years ago

Read 2 more answers

Calculating ph how is ph related to the concentration of hydronium ions. True or False

Dmitry_Shevchenko [17]

I think this is TRUE. Ph is calculating the acidic acids in water. And you need to know the concentration of hydronium ions. Hope this helped !

7 0

3 years ago