Calculate the speed of a car that travels 250 miles in 4.0 hours. (remember your unit)

Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge

Vanyuwa [196]

Answer:

a) F_net = 30.47 N , θ = 10.6º

b) Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

F₁₂ = k $k \frac{ q_{1} \ q_{2} }{ r^{2} }$

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

Fₓ = $F_{bc x}$

Y axis

$F_{y}$ Fy = $F_{ab} - F_{bc y}$

let's find the magnitude of each force

$F_{ab}$ = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

F_{ab} = 2.82 10¹ N

F_{ab} = 28.2 N

$F_{bc}$ = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

F_{bc} = 3.75 10¹ N

F_{bc} = 37.5 N

let's use trigonometry to decompose this force

tan θ = y / x

θ = tan⁻¹ and x

θ= tan⁻¹ ¾

θ = 37º

let's break down the force

sin 37 = F_{bcy} / F_{bc}

F_{bcy} = F_{bc} sin 37

F_{bcy} = 37.5 sin 37

F_{bcy} = 22.57 N

cos 37 = F_{bcx} /F_{bc}

F_{bcx} = F_{bc} cos 37

F_{bcx} = 37.5 cos 37

F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

Fₓ = 29.95 N

Axis y

Fy = 28.2 -22.57

Fy = 5.63 N

we can give the result in two ways

a) F_net = Fₓ i ^ + $F_{y}$ j ^

F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

F_net = $\sqrt{ F_{x}^2 + F_{y}^2 }$

F_net = √(29.95² + 5.63²)

F_net = 30.47 N

we use trigonometry for the direction

tan θ= $\frac{ F_{y} }{ F_{x} }$

θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }

θ = tan⁻¹ (5.63 / 29.95)

θ = 10.6º

3 0

3 years ago

White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w

REY [17]

Answer:

0.7515875 eV

$4\times 10^{14}\leq f$

Explanation:

f = Maximum frequency = $7.9\times 10^{14}\ Hz$

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

W = Work function = 2.52 eV

Converting to Joules

$W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J$

Maximum photon energy is given by

$E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J$

Maximum Kinetic energy is given by

$K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J$

Converting to eV

$1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV$

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

$W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz$

The range of frequencies for which no electrons are ejected is

$4\times 10^{14}\leq f$

5 0

3 years ago

OMG PLZ PLZ HELP!!

horsena [70]

1,3,4 I’m pretty sur e

4 0

3 years ago

Read 2 more answers

Find the length of an arc with a radius of 6.0m swept across 2.5 radians.

soldier1979 [14.2K]

Hello!

In terms of an arc length, radians is the measurement of an arc length in terms of the original radius. This arc length is 2.5 radians, so we multiply it by our original radius.

2.5(6)=15

Therefore, the arc length is 15 meters.

I hope this helps!

7 0

4 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$