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3 years ago

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A 2 kg blue car is moving 6 m/s to the right and collides with a 3 kg red car that is moving 2 m/s to the left. The cars collide

and stick together. What is their velocity after the collision? Include units on your answer.

1 answer:

snow_lady [41]3 years ago

7 0

Answer:

Their velocity after the collision is 1.2 m/s, to the right.

Explanation:

Given;

mass of the blue car, m₁ = 2 kg

initial velocity of the blue car, u₁ = 6 m/s

mass of the red car, m₂ = 3 kg

initial velocity of the red car, u₂ = 2 m/s

let the blue car moving to the right be in positive direction

also, let the red car moving to the left be in negative direction

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ - m₂u₂ = v(m₁ + m₂)

where;

v is their velocity after the collision

(2 x 6) - (3 x 2) = v(2 + 3)

12 - 6 = 5v

6 = 5v

v = 6/5

v = 1.2 m/s, to the right

Therefore, their velocity after the collision is 1.2 m/s, to the right.

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hichkok12 [17]

Answer:

a) $\Delta U_g=-5.3kJ$

b) $K=0.27kJ$

c) $F_f=0.45kN$

Explanation:

the gravitational potential energy is given by:

$U_g=m.g.h\\$

$\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ$

The kinetic energy is given by:

$K=\frac{1}{2}m.v^2\\$

the initial kinetic energy is zero because the motion started from rest, so:

$K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ$

applying the conservation of energy theorem:

$U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ$

The work done by the friction force is given by:

$W_f=F_f.h.cos(\theta)\\$

the angle of the force is 180 degrees because it's against the movement:

$F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN$

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Answer:

B) 3.6x 10_6 N/C or D)2.8 x105 N/C

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The angle of reflection is the angle the ____ to the reflecting surface.

Arlecino [84]

Answer:

The angle of reflection is the angle the reflected rays make with a perpendicular line to the reflecting surface.

Explanation:

Reflection It is the change of direction suffered by a luminous ray when hitting the surface of an object. The angle of reflection is that which is formed by the reflected ray and the normal vector to the study surface

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A car is traveling in uniform circular motion on a section of flat roadway whose radius is . The road is slippery, and the car i

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Answer:

The smallest radius will be four (4) times the initial radius

Explanation:

The car maintains a constant angular speed. According to Newton's Second Law F = m a

1. $F_{r}=m*A_{n}$

2. $A_{n}=\frac{v^2}{R_{p}}$

Replacing 2 in 1

3. $F_{r}=m*\frac{v^2}{R_{p}}$

Where:

Fr= Frictional force

Rp= Initial Radius

An= Centripetal Acceleration

M= Mass

V= Velocity

Also we have that:

4. $F_{r}=\mu *W=\mu*m*g$

μ= Coefficient of friction between the car and the surface

M= Mass

W= Weight

G= Gravity

r is cleared from equation 3

5. $R_{p}=m*\frac{v^2}{F_{r}}$

Replacing 4 in 5

6. $R_{p}=m*\frac{v^2}{\mu*m*g}$

Simplifying

7. $R_{p}=\frac{v^2}{\mu*g}$

Now we have a new velocity equal to twice the initial velocity, We replace it by 2v in equation 7

8. $R_{n}=\frac{(2v)^2}{\mu*g}$

Computing

9. $R_{n}=\frac{4v^2}{\mu*g}$

Replacing 5 in 9

$R_{n}=4*R_{p}$

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A rock is thrown horizontally from a bridge with a velocity of 17.0 m/s. It takes the rock 3.0 s to strike the water below. What

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Answer:

V= 33.98 m/s

Explanation:

Given that

Horizontal speed ,u= 17 m/s

Time taken by rockets to strike the water ,t= 3 s

We know that acceleration due to gravity ,g= 9.81 m/s²

There is no any acceleration in the horizontal direction that is why the horizontal veloity will remain constant.

In the vertical direction

vy = uy+ g t

Initial velocity in vertical direction is 0 m/s.

vy= 0+ 9.81 x 3

vy = 29.43 m/s

The resultant velocity

$V=\sqrt{v_y^2+u^2}$

$V=\sqrt{29.43^2+17^2}\ m/s$

V= 33.98 m/s

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