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nikdorinn [45]
3 years ago
5

A car heading north collides at an intersection with a truck of the same mass as the car heading east. If they lock together and

travel at 28 m/s at 37° north of east just after the collision, how fast was the car initially traveling? Assume that any other unbalanced forces are negligible.
what is the correct answer
34 m/s
17 m/s
26 m/s
68 m/s
...?
Physics
1 answer:
stealth61 [152]3 years ago
3 0

This phenomenon is an inelastic collision. The problem is easier to solve because the masses of the colliding bodies are equal. By conservation of momentum, the final momentum of the bodies as one is equal to the VECTOR sum of the momentum of individual bodies. 

For this problem, the final velocity of the two cars is the sum of the initial velocities of each car. The velocity of the car is just the component of the velocity along the north direction.

Using trigonometry,
v1 = vf * sin(theta)
v1 = 28 * sin(37)

v1 = 16.9 m/s
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Arunning back practices for his upcoming football game by running drills. He runs forward 62.4 yards, and runs backward for
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Answer: He has only move 0.2 yards

Explanation: When you subtract 18.3 from 18.5 you get 0.2 and that is how much he's moved

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3 years ago
A star emits electromagnetic radiation which closely resembles the spectrum of a blackbody. The three star system named Albireo
aksik [14]

Answer:

A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm , this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

we have a fluctuation of the intensity emitted by the stars.   Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

Explanation:

The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.

                  λ T = 2,898 10⁻³

therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm

The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

The burst with A = 4300K ​​has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.

Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

7 0
3 years ago
The instrument used to measure the diameter of a thin wire is​
Pie

micrometer is used to measure the diameter of a thin wire

3 0
4 years ago
Read 2 more answers
A. Draw the electric field lines around a negative charge.
Alborosie
<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric  fields, so this helps us to see the problem more real. So an electric field line is an imaginary  line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional  to the product of the charges and inversely proportional to the square  of the distance between them. </em>This can be expressed as follows:

F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m

Then:

F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge  and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

q_{1} \ and \ q_{2}

If the amount of charge on one of the objects is tripled, let's say this is the charge q_{2}, then the new charge is:

q_{N}=3q_{2}

In the formula of Coulomb:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2}

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod  repels positive charges on the sphere, creating  zones of negative and  positive charge as indicated in the second Figure.

7 0
3 years ago
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The sun doesn't set during what season and at what latitude?
Dahasolnce [82]

Answer: summer and greater than 66°N

Explanation:

Summer is the most appropriate season in which the sun does not set. It is a typical situation for the Arctic Circle. It is the latitude in which the sun does not set in the day and typically during the month of July. The north of this circle is a period of sunshine which last for a period of about six months in the North Pole. Summer never rises in the winter solstice (December).

4 0
3 years ago
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