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mars1129 [50]
3 years ago
6

A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o

f the initial acceleration vector and compare the initial acceleration vector’s magnitude with respect to the acceleration magnitude at the moment of the crash. [*** DRAW A DIAGRAM ***]
Physics
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

the direction of acceleration of the vehicle is the same direction of its velocity of car

s acceleration has the opposite direction to the car speed.

Explanation:

The initial acceleration of the car can be calculated with

          v = v₀ + a t

          a = (v-v₀) t

       

indicate that the initial velocity is zero (v₀ = 0 m / s)

         a = v / t

         a = 300 / t

the direction of acceleration of the vehicle is the same direction of its acceleration movement.

When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small

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A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

<u>F = 51.3°</u>

8 0
3 years ago
Which of the following BEST demonstrates a change in weight? *
Radda [10]

Answer:

A person changes from 100 kg to 75 kg after dieting

Explanation:

3 0
3 years ago
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Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If
frez [133]

Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2  

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2  

v2 = v1×A1/A2  

and  

r2 = 2×r1

then:

A2 = 4×A1  

so,

v2 = (v1)/4  

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2  

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)  

     = 9.75×10^4 Pa

    = 9.8×10^4 Pa

Therefore, the pressure in the wider section is 9.8×10^4 Pa

5 0
3 years ago
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the i
ELEN [110]

Answer:

Pls see attached file

Explanation:

5 0
3 years ago
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed
gregori [183]

Answer:

the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

Explanation:

Given the data in the question;

Let x represent the man's distance from building,

initially x = 1m2

dx/d t= -2.3 m/s

Also Let y represent shadow height

so we determine dy/dt when x is 4m from the building

form the image description of the problem, we see two-like triangles with the same base and height ratios

so

 2 / (12-x) = y / 12

24 = y(12 - x )

y = 24 / (12-x)

dy/dt = 24/(12-x)² × dx/dt

Now at x = 4,

we substitute

dy/dt will be;

⇒ 24/(12 - 4)² × -2.3

= 24/64 - 2.3

= 0.375 × -2.3

dy/dt = - 0.8625 m/s

Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

5 0
3 years ago
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