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WINSTONCH [101]
3 years ago
10

A 1000 kg red car traveling at 40 m/s rear ends a 3000 kg blue car traveling at 35 m/s. the cars bounce off each other and the b

lue car is accelerated to 37.5 m/s. what is the velocity of the red car after the crash?
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

32.5 ms⁻¹

Explanation:

<em><u>Theory </u></em>

<u> The law of conservation of linear momentum </u>

The sum of linear momentum of a system under no external force remains constant.

In this scenario although different forces act on two vehicles by each other when we  consider the total system of two cars no external unbalance force act on them. So we can apply this law.

As law says,

The sum of linear momentum before collision = The sum of linear momentum after collision.

Momentum = mass × velocity

1000×40+3000×35 = 1000×v + 3000×37.5

where v = velocity of the red car after the cash

v = 32.5 ms⁻¹

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Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof
il63 [147K]

<h2> The potential and kinetic energy of airplane are affected by these factors </h2>

Explanation:

When airplane rises up , it requires potential energy . This potential energy can be taken from the kinetic energy of airplane .

Thus if the speed of wind is larger , it can either oppose the motion of velocity or can favour the velocity of airplane  . By which its kinetic energy is effected .

If the weight of airplane is changed , it will effect the potential energy required . Thus heavier plane requires higher potential energy for attaining the same height .

Thus these two factor has important role in the flight of airplane .

6 0
3 years ago
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 pis
Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0
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By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged
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<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then: 

F = G(Mm / (4(pi)*r^2)) 

The above expression gives the force that you feel on the earth's surface, as it is today! 

Let us now double the mass of the earth and decrease its diameter to half its original size. 

This is the same as replacing M with 2M and r with r/2. 

Now the gravitational force (F' ) on the new earth's surface is given by: 

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F 

So: 

F' = 8F 

This implies that the force that you would feel pulling you down (your weight) would increase by 800%! 

You would be 8 times heavier on this "new" earth!</span>
4 0
3 years ago
What are valence electrons?
SashulF [63]

Answer:

The correct answer is D. Electrons in an atom that can bond with other atoms.

Explanation:

For those of you that need it still

8 0
2 years ago
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