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mixas84 [53]
4 years ago
12

Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

laced a total of 16 m, how long did it travel?
Physics
1 answer:
Luda [366]4 years ago
5 0

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*+\frac{1}{2} *2*8^{2}

x_{f}=-48+64\\x_{f}=16

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A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?
irakobra [83]

Answer:

10.8s

Explanation:

Given parameters:

Force on the car  = 3250N

Distance  = 35m

Power  = 11375W

Unknown:

Time taken = ?

Solution:

To solve this problem;

 Power is the rate at which work is done

         Power = \frac{work done }{time}  

  Work done  = force x distance  = 3250 x 35  = 123200J

Now;

          11375  = \frac{123200}{t}  

           11375t  = 123200  

                   t  = 10.8s

5 0
3 years ago
What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil
lora16 [44]

Answer:

\theta=145

Explanation:

The amplitude of he combined wave is:

B=2Acos(\theta/2)\\

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145

5 0
3 years ago
Durning which type of process does pressure remain consistent
Simora [160]

Answer:

\fbox {D. Isobaric}

Explanation:

The process during which pressure remains constant is called an isobaric process.

6 0
2 years ago
Read 2 more answers
A brick of mass 2.0kg is at rest. It falls to the ground through a
Lisa [10]

Answer:

I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

I hope my little bit (big you may say) hint help you with your question.

5 0
3 years ago
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
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