How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0σC is fired at 21.0 i^ m/s straight toward a second

MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1 = (m1+m2)v

After substituting the values we get,

v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

Ki +Ui = Kf + Uf

Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

1/2m1v1^2i = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf

After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

rf = 3.64m

(c)

v1f = (m1-m2 / m1 + m2) v1i

v1f = -9i m/s

(d)

v2f = (2m1/ m1 +m2) v1i

After substituting the values,

v2f = 12i m/ s

Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

To learn more about momentum conservation law click on the link below:

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5 0

2 years ago

Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13

lyudmila [28]

Answer:

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

$y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right )$

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

$y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right )$

$y = 0.14 Cos\left ( 2.512x-34.66t \right )$

3 0

3 years ago

Select the correct answer from each drop-down menu.

earnstyle [38]

Answer:

a transverse (sort of a plot of a sine or cosine graph, basically)

b longitudinal

c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)

Explanation:

7 0

3 years ago

A person paddles down river at an average speed of 4km

adoni [48]

Nice paddling. Thanks for sharing.
Do you have some question to ask ?

7 0

3 years ago

Read 2 more answers

Leoni is participating in four drawing competitions. If the probability of her losing any

Alex17521 [72]

Answer:

(a) $Probability = 0.7599$

(b) $Probability = 0.2646$

Explanation:

Represent losing with L and winning with W.

So:

$L = 0.7$ --- Given

$n = 4$

Probability of winning would be:

$W = 1 - L$

$W = 1 - 0.7$

$W = 0.3$

The question illustrates binomial probability and will be solved using the following binomial expansion;

$(L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

So:

Solving (a): Winning at least 1

We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4

So, we have:

$Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

Substitute value for W and L

$Probability = 4 * 0.7^3*0.3 + 6*0.7^2*0.3^2 + 4*0.7*0.3^3 + 0.3^4$

$Probability = 0.7599$

<em>Hence, the probability of her winning at least one is 0.7599</em>

Solving (a): Wining exactly 2

We look at the above and we list out the terms where the powers of W is exactly 2

So, we have:

$Probability = 6L^2W^2$

Substitute value for W and L

$Probability = 6*0.7^2*0.3^2$

$Probability = 0.2646$

<em>Hence, the probability of her winning exactly two is 0.2646</em>

6 0

3 years ago