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Mariulka [41]
3 years ago
11

A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one

wheel begins to stick. The train comes to a stop 7.7 m from the point at which it was released.
What is the train's acceleration after its wheel begins to stick?
Physics
2 answers:
Fittoniya [83]3 years ago
8 0
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
marta [7]3 years ago
6 0

Answer:

a = -1.16m/s^2

Explanation:

Given

Initial speed u = 2.5 m/s

Final speed v = 0

Distance traveled S = 7.7 m

Duration t = 2.0 s

Solution

Distance traveled before the wheel got stuck

d = ut\\\\d= 2.5 \times 2\\\\d = 5 m

Distance traveled after wheel got stuck

S'  = S - d\\\\S' = 7.7-5\\\\S' = 2.7 m

Acceleration

v^2 = u^2 + 2as'\\\\0 = 2.5^2 + 2 \times a times 2.7\\\\a = -1.16m/s^2

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Answer:

I = 0.11 A

Explanation:

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       P = V*I

  • In this case, the power is the one consumed by the cell phone = 390 mW, and the voltage the one produced by the internal energy of the battery, 3.5 V, neglecting the voltage loss at the internal resistance of the battery.
  • So, we can solve the above equation for  the current I, as follows:

        I = \frac{P}{V} = \frac{0.39W}{3.5V}  = 0.11 A

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3 years ago
A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

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An electric fan has the power output of 60W. How much work is done if the fan operates for 120s?
Digiron [165]
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4 0
3 years ago
a 45 kg ice skater initially skating at a velocity of 3 m/s speeds up to a velocity of 5 m/s. calculate the difference in the ma
ad-work [718]

Answer: 90 kgm/s

Explanation:

The momentum (linear momentum) p is given by the following equation:

p=m.V

Where:

m=45 kg is the mass of the skater

V is the velocity

In this situation the skater has two values of momentum:

Initial momentum: p_{1}=m.V_{1}

Final momentum: p_{2}=m.V_{2}

Where:

V_{1}=3 m/s

V_{1}=5 m/s

So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

p=p_{2}-p_{1}=m.V_{2}-m.V_{1}

p=m(V_{2}-V_{1})

p=45 kg(5 m/s - 3 m/s)

Finally:

p=90 kgm/s

4 0
3 years ago
Read 2 more answers
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
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