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olga_2 [115]
3 years ago
14

How do I find out where two cars intersect when I am given velocity for the red car 14.41cm/s and the starting point for the red

car 0.6m. I am also given velocity for the blue car 41.32cm/s and the starting point at 0m.
Physics
1 answer:
zimovet [89]3 years ago
3 0

Explanation:

Write equations for the position of each car as a function of time.  (Make sure to convert units if they're different.)

The red car's position x at time t is:

x = (14.41 cm/s) t + 60 cm

The blue car's position x at time t is:

x = (41.32 cm/s) t

To find when they meet, set the equations equal to each other.

(14.41 cm/s) t + 60 cm = (41.32 cm/s) t

60 cm = (26.91 cm/s) t

t = 2.230 s

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Physics !!!!!!!!!!!!!!!!!
madreJ [45]

Answer:

1) v = -0.89 m/s

2) v = 36.5 m/s

Explanation:

1) Given

The mass of the boy, M = 45 Kg

The mass of the skateboard, m = 2 Kg

The initial velocity of the boy and skateboard is zero

According to the law of conservation of linear momentum

                              MV + mv = 0

                                           V = - mv/M

Substituting the given values in the above equation

                                          V = -2 x 20/ 45

                                             = -0.89 m/s

Hence, the velocity of the boy, v = -0.89 m/s

2) Given

The mass of the boy, M = 47 Kg

The mass of the skateboard, m = 8 Kg

The initial combined velocity of the boy and skateboard, u = 4.2 m/s

The boy jumped off from the skateboard with velocity, V = -1.3 m/s

According to the law of conservation of momentum,

                      MV + mv = (M + m)u

                                    v = [(M + m)u - MV] / m

Substituting the given values,

                                     v = [(47 + 8)4.2 - 47(-1.3)] / 8

                                        = 36.5 m/s

Hence, the velocity of the skateboard, v = 36.5 m/s

3 0
4 years ago
A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the
WITCHER [35]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\

<u>v_f = 10.85 m/s</u>

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