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Tju [1.3M]
2 years ago
10

A good train took 7 hours to complete its journey. For the first 3 hours, it travelled an average speed of 186km/h. What was the

total distance of the journey? ____ km

Physics
1 answer:
Sliva [168]2 years ago
4 0
I really hope this helps

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Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a
wel

Answer:

a) v=7.32m/s

b) \alpha =-35º

c) ΔK=-1094.62J

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

p_{1x}=p_{2x}

To simplify the problem, lets name D for Daniel and R for Rebecca

a) p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}

Since Daniel's initial velocity is 0

m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}

v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}

v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s

Now, lets analyze the movement in the vertical direction

p_{1y}=p_{2y}

Since p_{1y}=0

0=m_{D}v_{D2y}+m_{R}v_{R2y}

v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s

Now, we can find the magnitude of Daniel's velocity after de collision

v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35º

c) The change in the total kinetic energy is:

ΔK=K_{2}-K_{1}

ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
2 years ago
Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from
Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

  • Density of the ice, ρ = 917 kg/m³
  • Initial pressure, P₁ = 1.0 bar
  • Final pressure, P₂ = 2.0 bar
  • Temperature, T = - 10 C
  • Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

\Delta G = V(P_2-P_1) \\\\

where;

V is the volume of the ice

Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L

Change in pressure;

P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm

The change in the Gibb's free energy per mole (G);

\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G =  0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

3 0
2 years ago
The speed of the incoming water is 25 cm/s and the cross-sectional area of the hose carrying the water is 2.0 cm2. 1) Determine
Sindrei [870]

Answer:

The time taken is  t = 500 \ s

Explanation:

From the question are told that

  The speed of the incoming water is  v =  25 cm/s

  The cross-sectional area of the host pipe is  a = 2.0 cm^2

Generally the rate at which the container is been filled is mathematically represented as

         \r  V  =  v *  a

=>      \r  V  = 25 * 2

=>      \r  V  = 50\  cm^3/s

Generally 2.5  L  =  25000 cm^3

Generally the time taken is  mathematically represented as

      t = \frac{25000}{50}

=>   t = 500 \ s

6 0
2 years ago
ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach
Simora [160]

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

6 0
3 years ago
In this graph, calculate the speed of<br> segment A in m/s?
andrew11 [14]

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

5 0
3 years ago
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