2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?

Given an array of words representing your dictionary, you test words to see if it can be made into another word in dictionary. T

katen-ka-za [31]

Answer:

Detailed solution is given below:

8 0

2 years ago

A hand crank generator is used to power a light bulb. The person turning the crank uses 22 Newtons of input force to turn the cr

ikadub [295]

Answer:

1. The input power is 17.6 Watts

2. The output power is 0.7232 Watts

3.The efficiency of the hand crank generator is approximately 4.109%

Explanation:

The given parameters are;

The force with which the crank generator is turned = 22 Newtons

The rate at which the crank is turned = 8 revolutions per 3 seconds

The distance the hand moves during each revolution = 0.3 meters

The current recorded by the multimeter = 0.08 amps

The voltage recorded by the multimeter = 9.04 volts

1. Power = The rate of doing work = Work done/(Time taken to do the work) = Force × Distance/Time

∴ The power input of the hand crank generator = 22 × 0.3 × 8/3 = 17.6 Watts

2. The power output to the bulb, P, is given by the formula for electrical power as follows;

Power = Current, I × Voltage V

∴ P = 0.08 × 9.04 = 0.7232 Watts

3. Efficiency in percentage = Output/Input × 100

Therefore, the efficiency of the hand crank generator = )Output power/(Input power)) × 100 = (0.7232/17.6) × 100 ≈ 4.109%

8 0

2 years ago

A paint company produces glow in the dark paint with an advertised glow time of 15 min. A painter is interested in finding out i

andrew11 [14]

Answer:

$p_v = P(z$

Now we can decide based on the significance level $\alpha$ . If $p_v$ we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

$\alpha=0.05$ we see that $p_v< \alpha$ so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

$\alpha=0.01$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

$\alpha=0.001$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

Explanation:

For this case they conduct the following system of hypothesis for the ture mean of interest:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu >15$

The statistic for this hypothesis is:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And on this case the value is given $z = -2.30$

For this case in order to take a decision based on the significance level we need to calculate the p value first.

Since we have a lower tailed test the p value would be:

$p_v = P(z$

Now we can decide based on the significance level $\alpha$ . If $p_v$ we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

$\alpha=0.05$ we see that $p_v< \alpha$ so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

$\alpha=0.01$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

$\alpha=0.001$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

3 0

2 years ago

: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the

Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}$

$V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s$

$\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}$

taking f = 0.0185

at Z₁ = Z₂

$\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}$

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P= 1.13\ psi$

c) When flow is downhill z₂-z₁ = -2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P=-0.601\ psi$

7 0

2 years ago

Assume that an attacker knows that a user’s password is either p1 = abcd or p2 = bedg. Say the user encrypts his password using

denis23 [38]

Answer: I will Show how the attacker can determine the user’s password

Explanation:

If we assume an attacker knows that a user’s password is either abcd or bedg. Say the user encrypts his password using the shift cipher, and the attacker sees the resulting ciphertext. Show how the attacker can determine the user’s password, or explain why this is not possible.

The alphabet{A, B, . . . , Z}is identified with the set Σ = {0, 1, . . . 25} and all additions are implicitly taken mod26.

Then, the possible passwords are p0 = abcd = (0, 1, 2, 3) and p1 = bedg = (1, 4, 3, 6).

All possible encryptions of p0 are C0 = {(k, k + 1, k + 2, k + 3) | k ∈ Σ} and the ones of p1 are C1 = {(k + 1, k + 4, k + 3, k + 6) | k ∈ Σ}.

These two sets are disjoint and so in order to obtain the password it is necessary to check in which set the ciphertext lies.

8 0

2 years ago