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GarryVolchara [31]
3 years ago
11

A car drives horizontally off the edge of a cliff that is 40.8 m high. the police at the scene of the accident note that the poi

nt of impact is 121 m from the base of the cliff. how fast was the car traveling when it drove off the cliff?
Physics
1 answer:
Inga [223]3 years ago
7 0
Fall time machine:
t = √ (2*H/g) = √ (2*40.8 / 9,8) ≈ 2.9 s
Horizontal movement:
S = V₀*t
Speed:
V₀ = S / t = 121 / 2.9 ≈ 41.7 m/s         or  150 km/h

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A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
kap26 [50]

Answer:

a) \omega \approx 219.911\,\frac{rad}{s}, b) \alpha = 16.916\,\frac{rad}{s^{2}}, c) a_{t} = 1.776\,\frac{m}{s^{2}}, d) a_{n} = 5077.889\,\frac{m}{s^{2}}, e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) v = 23.091\,\frac{m}{s}.

Explanation:

a) The maximum angular velocity of the fan is:

\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

\omega \approx 219.911\,\frac{rad}{s}

b) The angular acceleration of the fan is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}

\alpha = 16.916\,\frac{rad}{s^{2}}

c) The magnitude of the tangential aceleration is:

a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)

a_{t} = 1.776\,\frac{m}{s^{2}}

d) The magnitude of the centripetal acceleration is:

a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)

a_{n} = 5077.889\,\frac{m}{s^{2}}

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)

v = 23.091\,\frac{m}{s}

5 0
3 years ago
What is a type of pulley that increases the size and effort of force
12345 [234]
The answer is Movable
5 0
4 years ago
A spacecraft with a proper length of 250 m passes by an observer on the earth. according to this observer, it takes 0.770 µs for
Sophie [7]
If 1 second = 10000 μs
what about how many seconds = 0.77 μs
the calculations would be done lke this 
          0.77 x 1/1000000 = 7.7 x 10 ^ -7
Speed = distance / time
distance = length of the spacecraft
therefore speed = 250/7.7 x 10^ -7
ans = 324675324.7 m/s


4 0
3 years ago
MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle
labwork [276]

Answer:

(a)  y_{max}=0.423m

(b)  \alpha =64.3^{o}

Explanation:

Given data

v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle

Solution

For Part (a)

As the velocity component in direction of y is given by:

v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s

The maximum displacement is given by:

v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m

For Part (b)

To reach y=46cm =0.46m apply:

0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha  =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}

5 0
3 years ago
Read 2 more answers
Define force and provide an example​
ollegr [7]

Answer:

force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.

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4 years ago
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