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
belka [17]
3 years ago
6

An object moves along the x-axis according to the equation x = 3.00t2 – 2.00t + 3.00,

Physics
2 answers:
Ray Of Light [21]3 years ago
7 0

Explanation:

x = 3.00t^{2} – 2.00t + 3.00,

Distance of object at 2 second,

x (t=2) = 3(4) - 2(2) +3

x (t=2) = 12-4 +3

x (t=2) = 11 m

Distance of object at 3 second,

x (t=3) = 3(9) - 2(3) +3

x (t=2) = 27 - 6 + 3

x (t=2) = 24 m

a) the average speed between t = 2.00 s and t = 3.00 s,

Average speed = \frac{Total distance}{ Total time}

Average speed = \frac{x (t=2) + x (t=3)}{3}

Average speed = \frac{24+11}{3}

Average speed = \frac{35}{3}

Average speed = 11.66 \frac{m}{s}

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = \frac{dx}{dt}

Instantaneous speed(v) = 6t - 2\left \{ {{t=2} \atop {t=3}} \right.

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 \frac{m}{s}

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = \frac{average velocity}{time}

average acceleration =  \frac{11.66}{3-2}

average acceleration = 11.66 \frac{m}{s^{2} }

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = \frac{dv}{dt}

instantaneous acceleration =6

instantaneous acceleration = 6 \frac{m}{s^{2} }

e) for x =0

0 = 3.00t^{2} – 2.00t + 3.00

a = 3, b=-2, c=3

t= \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}

t= \frac{2 \pm \sqrt{4 - 36} }{6}

t= \frac{2 \pm \sqrt{-32} }{6}

general solution of this equation gives imaginary value. Hence, the given object is not at rest.

Aleks [24]3 years ago
5 0

a) the average speed between t = 2.00 s and t = 3.00 s,   v = 11.66  m/s

b) the instantaneous speed at t = 2.00 s and t = 3.00 s, v = 6  m/s

c) the average acceleration between t = 2.00 s and t = 3.00 s,  v = 11.66  m/s

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.  v = 6  m/s

e) At what time is the object at rest? the given object is not at rest.

<h3>Explanation: </h3>

An object moves along the x-axis according to the equation

x = 3.00t^2 - 2.00t + 3.00

where x is in meters and t is in seconds. Determine :

a) the average speed between t = 2.00 s and t = 3.00 s,

The instantaneous speed or speed is the object speed at a certain instant of time

Average speed = \frac{total distance}{ total time} = \frac{x(t=2s)+x(t=3s)}{3} \\

=\frac{(3.00*2^{2} - 2.00*2 + 3.00)+(3.00*3^{2} - 2.00*3 + 3.00)}{3} \\ =\frac{24+11}{3} \\ =\frac{35}{3} = 11.6 m/s

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = \frac{dx}{dt} = 6t - 2

6t - 2 where t=2 and t=3

Instantaneous speed(v) = v(t=2) - v(t=3) = 18-2- (12-2) = 6 m/s

c) the average acceleration between t = 2.00 s and t = 3.00 s, and

The instantaneous acceleration, or acceleration is the limit of the average acceleration when the interval of time approaches 0

average acceleration= \frac{average velocity}{time} =  \frac{11.66}{3-2} =  11.66 m/s

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.

instantaneous acceleration = \frac{dv}{dt} = 6t

6t where t=2 and t=3

Instantaneous acceleration (a) a(t=2 to t=3) = -( 6(2) -  6(3) )= -12 + 18 = 6 m/s^2

e) At what time is the object at rest?

at rest t=0  for x =0

0 = 3.00t^{2} - 2.00t + 3.00 using abc rules,

a = 3, b=-2, c=3

t= \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}

t= \frac{2 \pm \sqrt{4 - 36} }{6}

t= \frac{2 \pm \sqrt{-32} }{6}

The given object is not at rest because the general solution of this equation gives imaginary value

Learn more about  the instantaneous speed brainly.com/question/11686662

#LearnWithBrainly

You might be interested in
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i
mars1129 [50]

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  E = 7500 \ N/C

Explanation:

From the question we are told that  

       The  radius of the inner circle is r_i  =  0.80  \ m

        The  radius of the outer circle is  r_o  =  1.20 \ m

       The  charge on the spherical shell q_n  =  -500nC  = -500*10^{-9} \ C

      The magnitude of the point charge at the center is  q_c =  + 300 nC  =  + 300 * 10^{-9} \ C

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 E =  \frac{k *  q_c   }{x^2}

substituting values  

                  E =  \frac{k *  q_c   }{x^2}

where  k is  the coulomb constant with value k = 9*10^{9}  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

     substituting values

                  E =  \frac{9*10^9  *  300 *10^{-9}}{0.6^2}

                 E = 7500 \ N/C

7 0
3 years ago
(e)
Hatshy [7]

Answer:

≈933.3kg/m^3

Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

5 0
3 years ago
before dancing on a smooth wooden floor, ballet dancers sometimes sometimes put a sticky powered called rosin on their shoes sol
Katena32 [7]

Okay, so they want to basically Increase their grip, and they are taking advantage of the force of friction

3 0
2 years ago
Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

3 0
3 years ago
which statement best describes what would happen if the number of wore coils in a electromagnavneg were increased
jekas [21]
Where are the following choices
4 0
3 years ago
Other questions:
  • Pitch describes the loudness of a sound. <br> True False?
    5·1 answer
  • Calculations made using Celsius or Fahrenheit will not work for gas law calculations
    11·1 answer
  • A spring operated dart gun fires 10 g darts. Arming the gun requires 185 N of force and results in the shortening of the spring
    15·1 answer
  • An element's atomic number is the​
    6·1 answer
  • 2 examples of balanced forces
    6·1 answer
  • How are objects in space able to “fall” into orbit?
    15·1 answer
  • What are the characteristics of a nebulae? (Select all that apply.)
    10·1 answer
  • this is an area where two tectonic plates move towards each other with one going under the other plate?
    11·1 answer
  • First Amendment
    15·2 answers
  • Is light a particle or a wave? Is metal a good heat shield? Is the reason that nothing can go faster than light because we have
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!