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oee [108]
3 years ago
14

A 1100 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a

constant acceleration of 5.0 m/s2. At the same instant a 2000 kg truck, traveling at a constant speed of 7.0 m/s, overtakes and passes the automobile. A)How far is the com of the automobile-truck system from the traffic light at t=3.0 s? B)What is the speed of the com then?
Physics
1 answer:
Viktor [21]3 years ago
6 0

Answer:

a ) 21 m b) 7 m/s

Explanation:

For this case, we can consider for our reference system, that point zero (this is, at the rest of the traffic signal) is when X₀ = 0, and direction of movement is from left to right (positive sign)

For the car:  

a= acceleration is constant = 5 m/s² m1 , mass = 1100 Kg , V = not constant

This is an uniformly accelerated rectilinear movement, and applicable formulas are:

V = V₀ + at , X = X₀ + V₀t + 1/2at²

For the truck:

V = speed is constant = 7 m/s , mass = 2000 Kg, a = 0

This is an uniform rectilinear movement, and the applicable formula is:

X = X₀ + Vt

a) At t = 3.0 sec, we can use the formula for the track, considering that X₀ = 0 (when it pass the rest of the traffic signal)

X = 0 + (7 m/s)x(3 s) = 21 m  

So, at 3 sec, truck will be at 21 m from the rest of traffic light and, as truck has a constant speed; at that exact second; truck and car will be together as a com, so both will be 21 m away from point zero

b) At the exact time (3 sec, or in distance words, 21 m) car and truck will be together as a com, so they will both have the exact speed, hence, speed of car at that point will be 7 m/s

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masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

V(t) = \epsilon. e^{-t.\frac{L}{R} }

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

0.8*19 = 19. e^{-4.\frac{22}{R} }

0.8 = e^{-\frac{88}{R} }

ln(0.8) = ln(e^{-\frac{88}{R} })

ln(0.8) = -\frac{88}{R}

R = -\frac{88}{ln(0.8)}

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In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0
3 years ago
Can A positively charged body attract another positively charged body​
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Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

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The Stronger the force that is exerted on an object,the ____ the change in velocity will be.
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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze
sp2606 [1]

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

m_1 = Mass of penny = 0.0025 kg

m_2 = Mass of nickel = 0.005 kg

u_1 = Initial Velocity of penny = 2.35 m/s

u_2 = Initial Velocity of nickel = 0 m/s

v_1 = Final Velocity of penny

v_2 = Final Velocity of nickel

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s

The final velocity of the penny is 0.78333 m/s in the opposite direction

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s

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We see that for the same weight, if gravity is less, then the amount of mass is greater, because they are inversely proportional. So we conclude that the answer is:

<h2>a 3-N bowl of ice cream on the moon </h2>
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