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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

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Calculate the pressure on a man’s foot when a woman who weighs 520 N steps on his foot with her heel which has an area of 0.001
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Answer:

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6 0
3 years ago
5 points
olga2289 [7]

Answer:

d. 5 ohms

Explanation:

For resistors in parallel, the equivalent resistance is found with:

1/Req = ∑(1/R)

1/R = 1/15 + 1/15 + 1/15

1/R = 3/15

R = 15/3

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8 0
3 years ago
When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g
Olegator [25]

Answer:

Your question was incomplete so here is the complete question and answer.

Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)

a) plain water

b) 5-7 percent glucose solution

c) Glucose polymer solution of 6-8 percent

d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Explanation:

Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.

3 0
3 years ago
a person was using a spanner to loosen a tight nut.Their mass was 50kg the spanner was 0.5m long i) what is the persons weight?
sineoko [7]

1) Weight of the person: 490 N

2) Maximum torque: 245 Nm

Explanation:

1)

The weight of a body is equal to the gravitational force exerted on the body; it is given by the equation

F=mg

where

m is the mass of the body

g is the acceleration due to gravity

For the person in this problem,

m = 50 kg is the mass

g=9.8 m/s^2 is the acceleration due to gravity on Earth's surface

Therefore, the weight of the person is

F=(50)(9.8)=490 N

2)

The turning force (also called torque) exerted by a force rotating an object is given by

\tau = Fd sin \theta

where

F is the magnitude of the force

d is the length of the arm (the distance between the force and the pivotal point)

\theta is the angle between the direction of the force and the arm

For the spanner in this problem,

F = 490 N is the force applied (the weight of the person)

d = 0.5 m is the arm (the length of the spanner)

The maximum torque is obtained when \theta=90^{\circ}, therefore it is:

\tau=(490)(0.5)(sin 90^{\circ})=245 N\cdot m

Learn more about weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

About torque:

brainly.com/question/5352966

#LearnwithBrainly

5 0
3 years ago
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