A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?

Answers

Answer 1

Answer:

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².

v² - u² = 2 a ∆x, where u and v are initial and final velocities, respectively; a is acceleration.

and ∆x is the distance traveled (because the skater moves in only one direction).

Thus, (5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s².

Thus, A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².

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Answer 2

Answer:

Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s²

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The elasticity of demand for gasoline has been estimated to be 2.0, and the standard error is 1.0. The upper and lower bounds on the 95 percent confidence interval for the elasticity of demand for gasoline are:

Answers

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_(\alpha/2)(\sigma)/(√(n))$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_(\alpha/2)=1.96$

The standard error for this case is given:

$SE =(\sigma)/(√(n))=1$

Now we have everything in order to replace into formula (1):

$2-1.96*11=0.04$

$2+1.96*1=3.96$

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =2$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Calculate the confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_(\alpha/2)(\sigma)/(√(n))$ (1)

The standard error for this case is given:

$SE =(\sigma)/(√(n))=1$

Now we have everything in order to replace into formula (1):

$2-1.96*11=0.04$

$2+1.96*1=3.96$

A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Answers

Answer:

her initial frequency is 445 Hz

Explanation:

Given;

initial beat frequency, $F_B$ = 5

observed frequency, F = 440 Hz

let the initial frequency = F₁

F₁ = F ± 5 Hz

F₁ = 440 Hz ± 5 Hz

F₁ = 435 or 445 Hz

This result obtained shows that her initial frequency can either be 435 Hz or 445 Hz

The last beat frequency will be used to determine the actual initial frequency.

F = v/λ

Frequency (F) is inversely proportional to wavelength. That is an increase in length will cause a proportional decrease in frequency.

This shows that the final frequency is smaller than the initial frequency because of the increase in length.

Initial frequency - frequency of tuning fork = 5 beat frequency

Reduced initial frequency - frequency of tuning fork = 3 beat frequency

Initial frequency = 5Hz + 440 Hz = 445 Hz

Final frequency (Reduced initial frequency) = 440 + 3 = 443 Hz

Check: 445 Hz - 440 Hz = 5 Hz

443 Hz - 440 Hz = 3 Hz

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^(2) +2^(2) }$ = $√(13)$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^(2) + 3^(2) }$ = $√(25)$ = 5cm = 0.05m

electric potential V = $(kq)/(r)$

change in potential ΔV = $V_(1) - V_(2)$

ΔV = $(2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) )$ , where $q_(1) = q_(2)=$ 2.00μC

ΔV = $2kq((1)/(r_(1)) - (1)/(r_(2) ))$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $((1)/(0.036) - (1)/(0.05) )$

ΔV= 2.789×10⁵

$(1)/(2)mv^(2)$ = ΔV × q₃

$(1)/(2)$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

Read more about point charges at;brainly.com/question/13914561

The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to

Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

$p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))$

$p = 19051.2\,(lbf)/(yd^(2))$

14.7 psi is equal to 19051.2 pounds per square yard.

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 59.1 59.1 cm ( 0.591 0.591 m) and the flow speed of the petroleum is 11.9 11.9 m/s. At the refinery, the petroleum flows at 5.29 5.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answers

Answer:

The volume flow rate is 3.27m³/s

Diameter at the refinery is 88.64cm

Explanation:

Given

At the wellhead

Pipes diameter, d2 = 59.1cm = 0.591m

Flow speed of petroleum f2 = 11.9m/s

At the refinery,

Pipes diameter, d1 = ? Unknown

Flow speed of petroleum, f1 = 5.29m/s

Calculating the volume flow rate of petroleum along the pipe.

Volume flow rate = Flow rate * Area along the pipe

V = 11.9 * πd²/4

V = 11.9 * 22/7 * 0.591²/4

V = 3.265778m³/s

The volume flow rate is 3.27m³/s -------- Approximated

Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...

Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends

This gives;

V1A1 = V1A2

V1*πd1²/4 = V2 * πd2²/4 ----------- Divide through by π/4

So, we are left with

V1d1² = V2d2²

5.29 * d1²= 11.9 * 59.1²

d1² = 11.9 * 59.1²/5.29

d1² = 7857.172

d1 = √7857.172

d1 = 88.6406904305240618

d1 = 88.64cm --------------- Approximated

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?