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kakasveta [241]
2 years ago
15

Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

de R/R, es negativa. ¿cuáles son los ángulos posibles de R, medidos a partir del eje x positivo?
Physics
1 answer:
Masteriza [31]2 years ago
8 0
If you can write this in English I can help
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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
2 years ago
What voltage battery would you need to send 2.5A of current through a light bulb of resistance 3.6 ohm
nadezda [96]
<h3><u>Given</u> :</h3>

Current flow light bulb = 2.5\sf{A}

Resistance of light bulb = 3.6Ω

<h3><u>To Find </u>:</h3>

We have to find voltage of battery.

<h3><u>Solution</u> :</h3>

➠ As per ohm's law, current flow through a conductor is directly proportional to the applied potential difference.

➝ V ∝ I

➝ <u>V = I × R</u>

Where, R is the resistance of conductor.

⇒ V = I × R

⇒ V = 2.5 × 3.6

⇒ <u>V = 9 volt</u>

8 0
2 years ago
Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0
3 years ago
A car engine supplies 2.0 x 103 joules of energy during the 10. seconds it takes to accelerate the car along a horizontal surfac
andrew11 [14]

Answer: 2. 2.0*10^2 W

Explanation:

Power = Work/Time

Power = (2.0*10^3) Joules/10 seconds

Power = 2.0*10^2 Watts

7 0
2 years ago
What energy transformation occurs when a skydiver first jumps from a plane?
SSSSS [86.1K]
Hi
I think the answer is:
GRAVITATIONAL POTENTIAL ENERGY TRANSFORMS INTO KINETIC ENERGY.
HOPE IT HELPS.
4 0
3 years ago
Read 2 more answers
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