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mafiozo [28]
2 years ago
10

Before Collision:

Physics
1 answer:
marissa [1.9K]2 years ago
6 0

Answer:

  4/3 m/s

Explanation:

Assuming momentum is conserved, the sum of products of mass and speed before the collision is the same as after:

  (2000 kg)(4 m/s) +(4000 kg)(0 m/s) = (2000 +4000 kg)(Vt)

  Vt = (8000 kg·m/s)/(6000 kg) = 4/3 m/s

The speed of the combined objects after the collision is 4/3 m/s.

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Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram
bagirrra123 [75]

Speed of car A is given as

v_a = 70 mph

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

v_a = 70 *\frac{1609}{3600} = 31.3 m/s

now its distance from Bambi is given as

d_a = 350 m

time taken by it to hit the Bambi

t = \frac{d}{v}

t = \frac{350}{31.3}

t = 11.2 s

Now other car is moving at speed 50 mph

so its speed in SI unit will be

v_b = 50* \frac{1609}{3600}

v_b = 22.35 m/s

now its distance from Bambi is given as

d_b = 590 feet

as we know that 1 feet = 0.3048 m

d_b = 590*0.3048 = 179.83 m

now the time to hit the other car is

t_2 = \frac{179.83}{22.35}

t_b = 8.05 s

So Car B will hit the Bambi first

7 0
3 years ago
A car engine produces a force of 2000N while moving a distance of 200m in a time of 10s. What is the power developed by the engi
andreyandreev [35.5K]
Power = work/time
Work = force * distance

(2000 N * 200 m) / 10 s = 40,000 Joules
3 0
3 years ago
When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
2 years ago
A 50n box is lifted 2 meters in 3 seconds.
anygoal [31]

F=G=50 N

L=F*d=50*2=100 J

is the work

P=L/t=100/3=33.33 W

4 0
2 years ago
Which statement is true according to Newton's first law of motion?
katrin2010 [14]
C) In the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

hope this helps and have a great day :)







8 0
3 years ago
Read 2 more answers
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