1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ratelena [41]
3 years ago
6

A car is traveling at 50 ft/s when the driver notices a stop sign 100 ft ahead and steps on the brake. Assuming that the deceler

ation due to braking is constant, what is the magnitude of the minimum deceleration for which the car would stop right at the stop sign?
Engineering
1 answer:
Naddika [18.5K]3 years ago
4 0

Answer:

minimum deceleration = 12.5 ft/s²

Explanation:

Data provided in the question:

Initial speed of the car, u  = 50 ft/s

Distance to cover, s = 100 ft

Final speed of the car = 0 ft/s             [ ∵ as the car stops ]

Now,

from the Newton's equation of motion

we have  

v² - u² = 2as

a is the acceleration

on substituting the respective values, we get

0² - 50² = 2 × a × 100

or

- 2500 = 200a

or

a = - 12.5 ft/s²

Here, the negative sign depicts the deceleration.

You might be interested in
Which of the following is a way to heat or cool a building without using electricity or another power source?
zzz [600]
I believe the answer is: A. Passive heating and cooling.
8 0
3 years ago
Read 2 more answers
A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
Nataliya [291]

Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

And we can estimate the calue of C_D = 1.2 from a figure.

And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

8 0
3 years ago
The first assembled product used for testing and validating the product concept is called a prototype. True or false
harina [27]

Answer:

True

Explanation:

a prototype is first produced to test for defects

7 0
3 years ago
The contents of a tank are to be mixed with a turbine impeller that has six flat blades. The diameter of the impeller is 3 m. If
lions [1.4K]

Answer:

P=3.31 hp (2.47 kW).

Explanation:

Solution

Curve A in Fig1. applies under the conditions of this problem.

S1 = Da / Dt ; S2 = E / Dt ; S3 = L / Da ; S4 = W / Da ; S5 = J / Dt and S6 = H / Dt

The above notations are with reference to the diagram below against the dimensions noted. The notations are valid for other examples following also.

32.2

Fig. 32.2 Dimension of turbine agitator

The Reynolds number is calculated. The quantities for substitution are, in consistent units,

D a =2⋅ft

n= 90/ 60 =1.5 r/s

μ = 12 x 6.72 x 10-4 = 8.06 x 10-3 lb/ft-s

ρ = 93.5 lb/ft3 g= 32.17 ft/s2

NRc = (( D a) 2 n ρ)/ μ = 2 2 ×1.5×93.5 8.06× 10 −3 =69,600

From curve A (Fig.1) , for NRc = 69,600 , N P = 5.8, and from Eq. P= N P × (n) 3 × ( D a )5 × ρ g c

The power P= 5.8×93.5× (1.5) 3 × (2) 5 / 32.17 =1821⋅ft−lb f/s requirement is 1821/550 = 3.31 hp (2.47 kW).

5 0
3 years ago
Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radiu
Fynjy0 [20]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
4 years ago
Other questions:
  • Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an
    5·2 answers
  • Pro-Cut Rotor Matching Systems provide a non-directional finish without performing additional swirl sanding — true or false?
    13·1 answer
  • While you are driving you see an animal standing or walking on the side of the road. You should..
    9·1 answer
  • A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower
    15·1 answer
  • The purpose of hazard lights is to _____.
    11·2 answers
  • One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
    12·1 answer
  • HELP PLEASE NEED ANSWERS NOW Upload your energy audit project that includes the labeled sketch you used, Information about kilow
    14·1 answer
  • Which physical quantity measures fraction of the input energy a machine actually concerts into output energy ?
    13·1 answer
  • Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a
    6·1 answer
  • The end of a cylindrical liquid cryogenic propellant tank in free space is to be protected from external (solar) radiation by pl
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!