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Lady bird [3.3K]
3 years ago
6

A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th

e car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900.-kg car to rest from an initial speed of 1.1 m/s.
Physics
1 answer:
Alex Ar [27]3 years ago
3 0

Answer:

 avriage force F = 2722.5 N

Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

      vf² = v₀² - 2 ax

The final carriage speed is zero (vf = 0)

      0 = v₀² - 2ax

      a = v₀² / 2x

      a = 1.1²/(2 0.200)

      a = 3.025 m / s²

      a = 3.0 m/s²

We calculate the average force

      F = ma

      F = 900 3,025

      F = 2722.5 N

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The moon's orbital plane is different from that of the earth's orbit around the sun due to which lunar and solar eclipses do not occur every month.

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2 years ago
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(c) If η = 60% and TC = 40°F, what is TH, in °F?
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A particle moves along a straight path through a displacement d = 2.5i + cj while a force F = 8.5i + -8.5j acts on it. The displ
Zanzabum

Answer:

Explanation:

Work is defined as the scalar product of force and distance

W=F•d

Given that

F = 8.5i + -8.5j. +×-=-

F=8.5i-8.5j

d = 2.5i + cj

If the work in the practice is zero, then W=0

therefore,

W=F•ds

0=F•ds

0=(8.5i -8.5j)•(2.5i + cj)

Note that

i.i=j.j=k.k=1

i.j=j.i=k.i=i.k=j.k=k.j=0

So applying this

0=(8.5i -8.5j)•(2.5i + cj)

0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)

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8.5c=21.25

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3 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

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Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

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