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22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)
2N
250N
5000N
50000N

Answers

Answer 1

Answer:

Answer:

50000N

Explanation:

Force = mass × acceleration

= 2500 × 20

= 50000N

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_(beat) = 0.99s$

Generally the frequency of the beat is

$f_(beat) = (1)/(t_(beat))$

Substituting values

$f_(beat) = (1)/(0.99)$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_(beat)$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((231.01)^2)/((230)^2)$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_(beat)$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((228.99)^2)/((230)^2)$

$T_2 = 0.99$ % lower than $T_1$

A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centripetal acceleration of the car? 391 mi/h2 40.0 mi/h2 5,000 mi/h2 48,800 mi/h2

Answers

$a_(centrip)= (v^2)/(R)= (125^2 )/(0.32)=48828 (mi)/(h^2)$
The last one is the closest

48,800 mi/h2 is the right answer

Diethyl ether has a H°vap of 29.1 kJ/mol and a vapor pressure of 0.703 atm at 25.0°C. What is its vapor pressure at 60.0°C?

Answers

Answer:

The vapor pressure at 60.0°C is 2.416 atm

Explanation:

To solve this problem, we use Clausius-Clapeyron equation

$ln(P_2)/(P_1) = (-\delta H)/(R)[(1)/(T_2)-(1)/(T_1)]= (\delta H)/(R)[(1)/(T_1)-(1)/(T_2)]$

where;

Initial pressure P₁ = 0.703 atm

Initial Temperature T₁ = 25+273 = 298K

Final temperature T₂ = 60+273 = 333K

Change in enthalpy of vaporization ΔH = 29.1 KJ/mol = 29100J/mol

R is Boltzman constant = 8.314 J/K.mol

$ln(P_2)/(P_1) = (29100)/(8.314)[(1)/(298)-(1)/(333)] =1.23449$

$(P_2)/(P_1) = e^(1.23449)$ ⇒ $(P_2)/(P_1) = 3.43663$

P₂ = P₁ (3.43663) = (0.703 atm)(3.43663) = 2.416 atm

P₂ = 2.416 atm

Therefore, the vapor pressure at 60.0°C is 2.416 atm.

Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s2. How far will the car travel in 12 seconds?

Answers

Same formula as the last question. x = vt + (1/2)at^2. In this case, v = 0, t = 12, and a = 2.0. Plug in the values and solve for x (which is change in position)x = (0)(12) + (1/2)(2.0)(12^2)x = (1/2)(2.0)(144)x = (1)(144)x = 144So the car will travel 144 meters in 12 seconds.

Final answer:

The car, accelerating at a constant rate of 2.0 m/s2 from rest, will travel a distance of 144 meters in 12 seconds.

Explanation:

The question pertains to the concept of motion in physics, specifically how distances travelled are influenced by an object's acceleration. The car is accelerating at a constant rate of 2.0 m/s2 from rest. It means that the initial velocity of the car is 0. We can use the formula of motion, s = ut + 0.5at2, where u is the initial velocity, a is the acceleration and t is the time.

In this case, u = 0 (as the car starts from rest), a = 2.0 m/s2 (constant acceleration) and t = 12 seconds. Substituting these values into the formula, we get:

s = 0*12 + 0.5*2*122

Therefore, the car will travel 144 meters in 12 seconds assuming it accelerates at a constant rate.

Learn more about Constant Acceleration here:

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A gate with a circular cross section is held closed by a lever 1 m long attached to a buoyant cylinder. The cylinder is 25 cm in diameter and weighs 200 N. The gate is attached to a horizontal shaft so it can pivot about its center. The liquid is water. The chain and lever attached to the gate have negligible weight. Find the length of the chain such that the gate is just on the verge of opening when the water depth above the gate hinge is 10 m.