Let's say you have a plot for Pendulum experiment. Let's assume g for this experiment was measured (from the slope of the plot) to be 9.78 [ms^-2]. The vertical intercept, however, is 0.021 [s^2]. What might this translate to for a measurement of the length offset systematic in all the length measurements? Length Offset =[mm] .011

Answers

Answer 1
Answer:

Answer:

The length is 5.2 mm.

Explanation:

Given that,

Time period T²= 0.021 s²

Gravity due to acceleration = 9.78 m/s²

We need to calculate the length

Using formula of time period of pendulum

T=2\pi\sqrt{(l)/(g)}

l=(gT^2)/(4\pi^2)

Where, l = length

g = acceleration due to gravity

T = time period

Put the value into the formula

l=(9.78*0.021)/(4*\pi^2)

l=0.0052\ m

l=5.2\ mm

Hence, The length is 5.2 mm.


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A truck traveling with an initial velocity of 44.1 m/s comes to a stop in 15.91 secs. What is theacceleration of the truck?

Answers

Answer:

a=-2.77 m/s^2

Explanation:

Assuming constant acceleration,

v=at + v_0

where v_0 is the initial velocity.

At rest, v=0, so

0=at+v_0

So solving the equation for a:

a=(-v_0)/t

Inserting the numbers yields

a=-2.77 m/s^2

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 33.1 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Answers

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle=33.1^(\circ)

Let u be the initial speed of ball

Range=(u^2\sin 2\theta )/(g)

167=(u^2\sin (66.2))/(9.8)

u^2=1788.71

u=√(1788.71)

u=42.29 m/s

The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Answers

Part a)

Equation of position with time is given as

x = (2.0 m/s)t + (-3.0 m/s2)t^2

since this equation is a quadratic equation

so it will be a parabolic graph between t = 0 to t = 1

part b)

at t = 0.45 s

x = 2* 0.45 - 3 * 0.45^2

x_1 = 0.2925 m

at t = 0.55 s

x = 2* 0.55 - 3*0.55^2

x_2 = 0.1925

now the displacement is given as

d = x_2 - x_1

d = 0.1925 - 0.2925 = -0.1 m

so the average velocity is given by

v = (d)/(t)

v = \frac{-0.1}{0.1) = -1 m/s

part c)

at t = 0.49 s

x = 2* 0.49 - 3 * 0.49^2

x_1 = 0.2597 m

at t = 0.51 s

x = 2* 0.51 - 3*0.51^2

x_2 = 0.2397 m

now the displacement is given as

d = x_2 - x_1

d = 0.2397 - 0.2597 = -0.02 m

so the average velocity is given by

v = (d)/(t)

v = \frac{-0.02}{0.02) = -1 m/s

The electric field just above the surface of the charged drum of a photocopying machine has a magnitude E of 2.5 × 105 N/C. What is the surface charge density on the drum, assuming that the drum is a conductor?

Answers

Answer:

Charge_(density)=2.2125*10^(-6)C/m^(2)

Explanation:

Given data

Electric Field E=2.5×10⁵ N/C

To find

Charge Density

Solution

From definition of charge density we know that:

Charge Density=Electric field×Permttivity

Where Permttivity ∈₀=8.85×10⁻¹²C²/N.m²

Charge_(density)=(2.5*10^(5)N/C)*(8.85*10^(-12)C^(2)/N.m^(2))\n Charge_(density)=2.2125*10^(-6)C/m^(2)

20 examples of scalar quantity​

Answers

Answer:

Length

Time

Mass

Temperature

Energy

Direct Current (DC)

Frequency

Volume

Speed

Amount of substance

Luminous Intensity

Density

Concentration

Refractive Index

Work

Pressure

Power

Charge

Electric Potential

Entropy

If you know the distance of an earthquake epicenter from three seismic stations, how can you find the exact location of the epicenter of the earthquake.

Answers

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.