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Setler79 [48]
4 years ago
8

An initially motionless test car is accelerated uniformly to 115 km/h in 8.83 seconds before striking a simulated deer. The car

is in contact with the faux fawn for 0.995 seconds, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
Physics
1 answer:
goblinko [34]4 years ago
3 0

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity before collision = 115 km/hr = 115*5/18 = 31.94 m/s

Time taken by car to reach this velocity = 8.83 seconds

Initial velocity = 0 m/s

v = u +at

31.94 = 0 + a*8.83

a = 3.62 m/s^2

So acceleration of car just before collision = 3.62 m/s^2

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Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d
lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

d = d_1 + d_2 + d_3

d = 7 km + 5 km + 1 km

d = 13 km

Now in order to find the displacement we will show all with their directions

\vec d_1 = 7 + 5 = 12 km towards North

\vec d_2 = 1 km towards East

So total displacement is

\vec d = \vec d_1 + \vec d_2

\vec d = 12 \hat j + 1 \hat i

so net displacement will be

d = \sqrt{12^2 + 1^2} = 12.04 km

so displacement is 12.04 km

6 0
4 years ago
A block of wood of mass 300g and density 0.75 g/cm^3 is floating on the surace of a liquid of density 1.1 g/cm^3. What mass of l
Travka [436]

Answer:

The minimum mass of the lead for the combination to submerge is 155 g.  

Explanation:

let M be the mass of the wood.

let m be the minimum mass of lead to be added for the combination to submerge.

let ρ1 be the density of the liquid.

let ρ2 be the density of the wood.

let ρ3 be the density of lead.

let g be the gravitational acceleration.

For the combination to submerge, the weight of the wood combined with the weight of lead should at least be equal to the buyant force, that is:

weight of wood and lead = buyant force

g×(M+m) = g×(ρ1)×(M/ρ2 + m/ρ3)

M+m = (ρ1)×(M/ρ2 + m/ρ3)

m - ρ1×m/ρ3 = (ρ1)×(M/ρ2) - M

m(1 - ρ1/ ρ3) = M(ρ1/ρ2 -1)

m = [M(ρ1/ρ2 -1)]/[(1 - ρ1/ ρ3)]

   = [(300)(1.1/0.75 -1)]/[(1 - 1.1/ 11.3)]

   = 155 g

Therefore, the minimum mass of the lead for the combination to submerge is 155 g.  

4 0
4 years ago
Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th
dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

         

Let's write the diffraction equation for a slit

       a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

             θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

           θ = 1.22 λ / D

Where D is the diameter of the pupil

 Let's apply this equation to our case

        θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

        θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0
3 years ago
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