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
quester [9]
3 years ago
13

A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s for 6.2 s. It then accelerates at a rate of-1.2

m/s until it stops. a)Find the car's maximum speed. b) Find the total time from the start of the first acceleration until the car is stopped. c) What is the total distance the car travelled?
Physics
2 answers:
gregori [183]3 years ago
6 0

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.26 s

c) Total distance traveled = 390,5537

Explanation:

In order to solve the first proble we just have to use the next formula:

Vf= Vo+ Acc-t\\Vf= 13,5 + 6,2*1,9\\Vf=25,28 m/s\\

So the maximum speed would be 25,28 m/s.

THe total time of the trip will be given by adding the inital time plus the velocity divided by the acceleration rate:

Time= 6,2 s+ \frac{25,28}{-1,2} \\Time= 6,2 +21,06\\Time= 27,26

Remember that when dealing with time in physics you will always use positive numbers since there is no negative time.

To calculate the total distance covered we use the next formula:

D=Vo*t+ 1/2a*t^2\\D= 13,5*6,2 + 1/2(1,9)(6,2)^2\\D=87,75+36,518\\D=124,268m

This is the first part now we calculate it with the stopping of the car:

D=Vo*t+ 1/2a*t^2\\D=25,28*21,06+ 1/2(-1,2)(21,06)^2\\D=532,3968-266,1141\\D= 266,2857meters

No we just add the two distances to discover the whole distance:

Total distance= 124,268+266,2857= 390,5537

yanalaym [24]3 years ago
5 0

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.27 s

c) Total distance traveled = 402.43 m

Explanation:

a) Maximum speed is obtained after the end of acceleration

        v = u + at

        v = 13.5 + 1.9 x 6.2 = 25.28 m/s

    Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

         v = u + at  

         0 = 25.28 - 1.2 t

         t = 21.07 s

    Total time = 6.2 + 21.07 = 27.27 s

c) Distance traveled for the first 6.2 s

          s = ut + 0.5 at²

          s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m

   Distance traveled for the second 21.07 s

          s = ut + 0.5 at²

          s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m

   Total distance traveled = 120.22 + 282.21 = 402.43 m

You might be interested in
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
The pickup truck has a changing velocity because the pickup truck
yaroslaw [1]

Answer:

travilng on a curve in the road

Explanation:

7 0
3 years ago
Read 2 more answers
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and
Lunna [17]

Answer:

Q = - 4312 W = - 4.312 KW

Explanation:

The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:

Q = - kA dt/dx

Integrating from one side of the slab to other along the thickness dimension, we get:

Q = - kA(T₂ - T₁)/L

Q = kA(T₁ - T₂)/t

where,

Q = Rate of Heat Loss = ?

k = thermal conductivity = 1.4 W/m.k

A = Surface Area = (11 m)(8 m) = 88 m²

T₁ = Temperature of Bottom Surface = 10°C

T₂ = Temperature of Top Surface = 17° C

t = Thickness of Slab = 0.2 m

Therefore,

Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m

<u>Q = - 4312 W = - 4.312 KW</u>

<u>Here, negative sign shows the loss of heat.</u>

3 0
3 years ago
Hi :) how to do this?
fredd [130]

Answer:

The answer is b

Hope this helps you!

4 0
3 years ago
Will mark brainliest.
egoroff_w [7]

Answer:

pls can you get a clearer image, then u can report back to me

6 0
3 years ago
Other questions:
  • your friend rides her bicycle across town at a constant speed. Describe how you could determine her speed
    12·1 answer
  • Oil _____.
    11·2 answers
  • 3. A clown rides a small car at a speed of
    15·1 answer
  • TESE
    10·1 answer
  • The ice in his drink has a mass of 6 g and a volume of 20 cm3. what is the density of the ice?
    10·1 answer
  • Fatigue strength is generally significantly improved by using high steel a. alloy b. yield c. hardened d. ultimate strength e. a
    14·1 answer
  • In any electromagnetic wave,
    13·1 answer
  • Convert Rev/min to rad/s x 2pie/60?<br><br> Anyone knows this please?
    11·1 answer
  • Please I really need the help
    13·1 answer
  • What is the phenomenon that can occur if the percentage of oxygen is higher in the air?
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!