The number of oscillations that a wave completes per unit of time in called its ______

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

3 years ago

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A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe

nataly862011 [7]

Answer:

$v_x = 11.78 m/s$

Explanation:

Velocity of the ship is given as

$v = 12 units$

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

$v_x = v cos\theta$

$v_x = 12 cos11$

$v_x = 11.78 m/s$

7 0

3 years ago

Which statement best summarizes the theory of evolution by natural selection?

prisoha [69]

Answer:

c is the one that makes the most sense

Explanation:

8 0

3 years ago

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A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

12345 [234]

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

v = √Rg

v = √(1.0(9.8))

v = 3.1304951...

v = 3.1 m/s

6 0

3 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at

devlian [24]

Force acting during collision is internal so momentum is conserve so (initial momentum = final momentum) in both directions Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west. Let Vx is and Vy are final velocities of car in +x and +y direction respectively. initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)

- 750*25 + 1150*0 = (750+1150)
Vx initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)

750*0 - 1150*5 = (750+1150)
Vy from here you can calculate Vx and Vy so final velocity V is

V=(√V2x+V2y)

and angle make from +ve x axis is

θ=tan−1(VyVx)

 kinetic energy loss in the collision = final KE - initial KE

5 0

3 years ago

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The number of oscillations that a wave completes per unit of time in called its _______.