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blondinia [14]
3 years ago
12

Two cars leave Calgary at the same time, travelling in opposite directions. Their average speeds differ by 5 km/h. After 2 h, th

ey are 210 km apart. Find the speed of each car.
Physics
2 answers:
dimulka [17.4K]3 years ago
8 0
Let the first car's average speed be x

The second car's speed would be x+5

2(x+(x+5))=210

2x+5=105

2x=100

x=50

So the speed of the slower car is 50mph, and the speed of the faster car is 55mph


DerKrebs [107]3 years ago
3 0
I think the answer is 50 and 55 but to check that u can multiply both of them by two and then add them together and see if it = 210
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Knowing Newton’s 2nd Law, how would you rearrange it to solve for acceleration.
AysviL [449]

Answer:

Force / mass

Explanation:

Divide mass on both sides to get acceleration by itself leaving you with mass below force hence divide force by mass

8 0
3 years ago
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A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), Fig. 5–39.
Trava [24]

Answer:

1742.24106 revolutions per day

Explanation:

v = Velocity

d = Diameter = 1.1 km

r = Radius = \dfrac{d}{2}=\dfrac{1.1}{2}=0.55\ km

g = Acceleration due to gravity = 9.81 m/s²

g = 0.9 g

The centrifugal force will balance the gravitational force

F_c=mg\\\Rightarrow \dfrac{mv^2}{r}=m0.9g\\\Rightarrow v=\sqrt{\dfrac{0.9gmr}{m}}\\\Rightarrow v=\sqrt{0.9gr}\\\Rightarrow v=\sqrt{0.9\times 9.81\times 0.55\times 10^3}\\\Rightarrow v=69.68464\ m/s

\dfrac{1}{T}=\dfrac{v}{2\pi r}\\\Rightarrow \dfrac{1}{T}=\dfrac{69.68464}{2\pi 0.55\times 10^3}\times 24\times 60\times 60\\\Rightarrow \dfrac{1}{T}=1742.24106\ rev/day

The rotation speed is 1742.24106 revolutions per day

6 0
3 years ago
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Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
Anni [7]

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass  ×  g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction  

Fa = 74.97 × 0.10

Fa =  7.49 N

and

force for moving constant velocity is

Fa =  weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

7 0
3 years ago
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Does anyone know what the answers are?
baherus [9]

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

4 0
3 years ago
Un casquillo de cobre de 8 kilogramos tiene que calentarse de 25 a 140 grados Celsius con el fin de expandirlo pata que se ajust
uranmaximum [27]

Answer:

354200J

Explanation:

Given parameters:

Mass of copper bushing = 8kg

Initial temperature  = 25°C

Final temperature  = 140°C

Unknown:

Quantity of heat required to heat this mass = ?

Solution:

The amount of heat required to heat mass from one temperature to another is given by;

                  H = m c Δt

where m is the mass

          c  is the specific heat

        Δt is the change in temperature

C is a constant and for copper, its value is 385J/kg°C

 Input the parameters;

            H = 8 x 385 x (140 - 25) = 354200J

8 0
3 years ago
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