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Rzqust [24]
3 years ago
9

Highway transportation officials are trying to determine if the average speed on Archer road is above 45mph, the post speed limi

t. They took a sample of 50 cars. The sample mean was 44.9 and the standard deviation was 10. Find the test statistic.0.010.0707-0.01-0.0707
Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

Option D) -0.0707

Explanation:

We are given the following in the question:  

Population mean, μ = 45 mp

Sample mean, \bar{x} = 44.9

Sample size, n = 50

Sample standard deviation, s = 10

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{44.9 - 45}{\frac{10}{\sqrt{50}} } = -0.0707

Thus, the test statistic is

Option D) -0.0707

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A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of gla
Diano4ka-milaya [45]

Complete Question

A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,  

A :

a force repels the glass out of the capacitor.  

B :

a force attracts the glass into the capacitor.    

C :

no force acts on the glass.      

D :

a net charge appears on the glass.      

E :

the glass makes the plates repel each other.

Answer:

The correct option is B

Explanation:

Generally when the glass dielectric is slowly inserted between the plated,

The positive plate of the capacitor will induce a negative charge on the glass while the negative  plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass

3 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

8 0
3 years ago
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F=ma
where:
F - force
m - mass
a - acceleration 

We transform this formula to get a:
a= \frac{F}{m}
a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
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3 years ago
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