A student's pencil rolls off a desk and lands 0.47 m away from the edge of the

edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement

Naily [24]

The displacement of Edward in the westerly direction is determined as 338.32 km.

<h3>What is displacement of Edward?</h3>

The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.

The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰

The displacement is the side of the triangle facing 150⁰ = R

R² = a² + b² - 2abcosR

R² = 150² + 200² - (2x 150 x 200)xcos(150)

R² = 62,500 - (-51,961.52)

R² = 114,461.52

R = 338.32 km

Learn more about displacement here: brainly.com/question/321442

#SPJ1

3 0

1 year ago

The model of the atom proposed by Greek philosophers appears similar to the model proposed centuries later by Dalton. What was t

Sladkaya [172]

The main difference between the model of the atom proposed by Greek philosophers and the model proposed centuries later by Dalton is that the Greek one was mainly speculative and philosophical - it wasn't based on real evidence, but on their suggestions and thoughts about the matter. On the other hand, Dalton had the means to prove his theory using viable evidence, not just speculations.

4 0

3 years ago

An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How

Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

$g=24 m/s^2$

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

$s=0+\frac{1}{2}(24)(1)^2=12 m$

6 0

3 years ago

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you

soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

$|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s$

The interaction time to avoid that the water balloon breaks is 0.029s

5 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago