PHYSICS HIGH SCHOOL

A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation x = 0.5 sin (pt+p/3). The acceleration (in m/s2) of the body at t = 1.0 s is approximatelya. 3.5
b. 49
c. 14
d. 43
e. 4.3

Answers

Answer 1
Answer: I'll tell you how I look at this, although I may be missing something important.

Position = x(t) = 0.5 sin(pt + p/3)

Speed = position' = x'(t) = 0.5 p cos(pt + p/3)

Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)

At (t = 1.0),

x ' '(t) = -0.5 p² sin( 4/3 p )

In order to evaluate this, don't I still have to know what 'p' is ? ?

I don't think it can be evaluated with the information given in the question.

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If the sun rises at 5 am and sets at 6 pm, what is the length of the day? a) 11 hours b) 12 hours c) 13 hours d) 24 hours

Answers

Answer:

24 hours

Explanation:

Within an internal combustion engine, the can-shaped component that moves up and down the cylinder and is the main moving part in the engine is theA. connecting rod.
B. piston.
C. spark plug.
D. crankshaft.

Answers

Piston is the part in the combustion engine which moves up and down. Piston is connected in connecting rod which helps to move it.
Spark plugs are used to produce spark.

So the answer is B. Piston

Hope I Helped You!!! :-)

Have A Great Day!!!
Piston is the part in the combustion engine which moves up and down.
So your answer is B.

Cell phones use digital signals to send and receive information. These digital signals travel as what kind of wave?ОООО
Gamma rays
X-rays
Microwaves
Radio waves

Answers

the answer would be microwaves!

Answer:

Microwaves

Explanation:

Explanation:

Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far from it would you place a sphere representing the earth? (Center to center distance please.) The distance from the center of the sun to the center of the earth is 1.496×10111.496×1011 m and the radius of the sun is 6.96×1086.96×108 m.

Answers

In this exercise we have to use the knowledge in distance, in this way we will find that the proportional distance found is:

d = 0.645 m

So from the information given in the text we find that:

  • The distance from the center of the sun to the center of the earth is 1.496*10^(11) \ m
  • The radius of the sun is 6.96*10^(8)m
  • We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm  

First, we need to find in what way or manner often the radius of the brightest star exist considerable respect to the range of the ball significance, that exist given apiece following equating:

(r_a)/(r_b)= \frac{6.96*10^8}{3*10{-3}} =2.32*10^(11)

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth:  

d_(s) = (d_(e))/(r_(s)/r_(b)) = (1.496 \cdot 10^(11) m)/(2.32\cdot 10^(11)) = 0.645 m

See more about distance at brainly.com/question/989117

Answer:

d = 0.645 m(assuming a radius of the ball bearing of 3 mm)

Explanation:

The given information is:

  • The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = d_(e)
  • The radius of the sun is 6.96x10⁸m = r_(s)

We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = r_(b).  

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

(r_(s))/(r_(b)) = (6.96\cdot 10^(8)m)/(3\cdot 10^(-3)m) = 2.32\cdot 10^(11)

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, d_(s):  

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

Two stars in the night sky make an angle of 10^-3 radians at the eye of Matilda. If Arthur, who can just distinguish two spots, 0.5 cm apart, at a distance of 10 m views the stars, can he distinguish them as two distinct objects?

Answers


Let's calculate the angle between two spots, 0.5 cm apart,
that are 10m  away from you.

1 m  =  100 cm
10m  =  1,000 cm

The angle between the two spots is very very close to the angle
whose tangent is
                              0.5cm/1,000cm  =  0.0005 .

That angle is   tan⁻¹(0.0005) =  0.0005 radian  =  0.029 degree .

Matilda is looking at two stars that appear to be  0.001  radian apart.

That's (0.001 / 0.0005) = 2 = TWICE as far apart as Arthur can distinguish.

As long as Arthur and Matilda are on the same planet while looking at
these two stars ... or as long as they're even in the same solar system ...
Arthur will have no trouble separating them.

Which type of heating system is often used to heat many buildings from a central location

Answers

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