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Tems11 [23]
3 years ago
7

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

stop but imparts all it’s mometum to the truck. How fast is the truck moving after the collision and in which direction? (Hint: this is an elastic collision)
Physics
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

<em><u>M</u></em><em><u>a</u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:</u></em>

That will be

<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>

<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>

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Which of the following methods has led to the earliest discoveries of massive planets orbiting near their parent stars a. detect
Mrac [35]

Answer:

c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum

Explanation:

In a solar system the mass of the star and planets affect each other's orbital  movements. The center of gravity of a star and a planet is inside the star. This causes the star to be closer and farther from the Earth at different times. Due to this wobble the star appears to be red shifted when it is farther and blue shifted when it is closer.

When the mass of the planet is high, like a hot Jupiter it causes more wobble i.e., change in radial velocity. This makes it easier to detect the planet. The earliest hot Jupiter found by this method is the planet 51 Pegasi b.

3 0
4 years ago
Marc and Linh stretch out a long spring on the classroom floor. Marc holds one end of the spring still. Linh creates waves in th
Soloha48 [4]

The speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

<h3>How to calculate speed of a wave?</h3>

The speed of a wave can be calculated by using the following formula:

Speed = Wavelength x Frequency

According to this question, Linh creates waves in the spring by moving the other end right and left with a frequency of 2 Hz. If wave crests are 0.5 m apart, the speed can be calculated as follows:

speed = 2Hz × 0.5m

speed = 1m/s

Therefore, the speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

Learn more about speed at: brainly.com/question/10715783

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7 0
2 years ago
Which formula can be used to find the magnitude of the resultant vector? R2 = Rx2 + Ry2 R = Rx + Ry R = Rx(cosθ) R = Rx(sinθ)
12345 [234]
R^2 = rx^2 + ry^2 !!!!!!!!!!
5 0
3 years ago
Read 2 more answers
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
4 years ago
Perpetual motion machines have fascinated people especially inventors for hundreds of years before the laws of thermodynamics be
Cerrena [4.2K]

Given the fact that energy conversion is not entirely efficient, it is impossible to produce a perpetual motion machine.

<h3>What is a perpetual motion machine?</h3>

The perpetual motion machine in one that is able to work continuously without stopping. This would mean that the efficiency of this machine must that the machine is 100% efficient which violates the second law of thermodynamics.

Thus, given the fact that energy conversion is not entirely efficient and energy looses cause machines not function effectively, it is impossible to produce a perpetual motion machine.

Learn kore about a perpetual motion machine:brainly.com/question/13001849

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5 0
2 years ago
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