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salantis [7]
3 years ago
7

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car's motion, the east

ward component of the car's velocity is given by _x(t)=(0.820m/s3)t2.
Physics
1 answer:
son4ous [18]3 years ago
7 0
The solution to your problem is as follows:

a=dv/dt=2*0.82*twhat time velocity is 13.7 solve 13.7=0.82*t^2   t=sqrt(13.7/0.82)  =4.51 sUse the time in the acceleration expression a= 2*0.82*4.51=7.39m/s^2


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Which steps can be taken to translate the phrase “the height of a tree is increased by seven inches” into an algebraic expressio
ololo11 [35]

Answer:

Explanation:

height of tree = x

seven inches= 7

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3 0
2 years ago
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A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

E = 10r, E = 10t

3 0
3 years ago
A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical
DiKsa [7]

Answer:

The correct answer is "64 J".

Explanation:

The given values are:

Mass,

m = 52 kg

Velocity,

v = 6 m/s

Mechanical energy,

= 1000 J

Now,

The gravitational potential energy will be:

⇒ P.E=1000-\frac{1}{2}mv^2

           =1000-\frac{1}{2}\times 52\times (6)^2

           =1000-26\times 36

           =1000-936

           =64 \ J

7 0
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A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg
Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0
3 years ago
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 7
Kisachek [45]

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2

Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s

Since, 1 m/s = 2.23 mph

4.28 m/s = 9.57 mph

So, the speed of the rider is 4.28 m/s or 9.57 mph.  

5 0
3 years ago
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