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The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

Answers

Answer 1

Answer:

(1) The acceleration of the car will be $a=-3.5(m)/(s^2)$

(2) The time taken $t=5.7s$

(3) The time is taken by the car to slow down from 20m/s to 10m/s $t=2.9s$

What will be the acceleration and time of the car?

(1) The acceleration of the car will be calculated as

$a=(v-u)/(t)$

Here

u= 14 $(m)/(s)$

$a=(0-14)/(4) =-3.5(m)/(s^2)$

(2) The time is taken for the same acceleration to 20 $(m)/(s)$

$a=(v-u)/(t)$

$t=(v-u)/(a)$

u=20 $(m)/(s)$

$t=(0-20)/(-3.5) =5.7s$

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

$t=(v-u)/(a)$

v=10 $(m)/(s)$

u=20 $(m)/(s)$

$t=(10-20)/(-3.5) =2.9s$

Thus

(1) The acceleration of the car will be $a=-3.5(m)/(s^2)$

(2) The time taken $t=5.7s$

(3) The time is taken by the car to slow down from 20m/s to 10m/s $t=2.9s$

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Answer 2

Answer:

(a) $-3.5 m/s^2$

The car's acceleration is given by

$a=(v-u)/(t)$

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

$a=(0-14)/(4)=-3.5 m/s^2$

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

$a=(v-u)/(t)$

where in this case we have

$a=-3.5 m/s^2$ is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

$t=(v-u)/(a)=(0-20)/(-3.5)=5.7 s$

(c) $2.9 s$

As before, we can use the equation

$a=(v-u)/(t)$

Here we have

$a=-3.5 m/s^2$ is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

$t=(v-u)/(a)=(10-20)/(-3.5)=2.9 s$

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A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?

Answers

Answer:

$f=1.79Hz$

Explanation:

The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.

$f=(1)/(T)\nf=(1)/(0.56 s)\nf=1.79Hz$

The speed of light is 3 x 10 m/s.Calculate the frequency of light that is absorbed the most by the 100m length of fibre.
Give your answer in standard form.

Answers

Answer:

$3 * 10 {}^6$

1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.

Answers

The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

Given data:

The mass of sphere is, m = 10 kg.

The radius of sphere is, r = 0.5 m.

The density of air is, $\rho = 1.22 \;\rm kg/m^(3)$ .

The drag coefficient of object is, $C_(d)=0.8$ .

The altitude is, h = 5000 m.

The density of air at altitude is, $\rho' =0.736 \;\rm kg/m^(3)$ .

The mathematical expression for the terminal velocity of an object is,

$v_(t)=\sqrt(2mg)/(\rho * A * C_(d))$

here,

g is the gravitational acceleration.

A is the area of sphere.

Solving as,

$v_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi r^(2)) * C_(d))}\n\n\nv_(t)=\sqrt{(2 * 10 * 9.8)/(1.22 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\n\v_(t)=7.99 \;\rm m/s$

Now, the terminal velocity at the altitude of 5000 m is given as,

$v_(t)'=\sqrt(2mg)/(\rho' * A * C_(d))$

Solving as,

$v_(t)'=\sqrt{(2 * 10 * 9.8)/(0.736 * (4 \pi * 0.5^(2)) * 0.8)}\n\n\nv_(t)'=10.30 \;\rm m/s$

Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.

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Answer:

The terminal velocity at sea level is 7.99 m/s

The terminal velocity at an altitude of 5000 m is 10.298 m/s

Explanation:

mass of sphere m = 10 kg

radius of sphere r = 0.5 m

air density at sea level p = 1.22 kg/m^3

drag coefficient Cd = 0.8

terminal velocity = ?

Area of the sphere A = $4\pi r^(2)$ = 4 x 3.142 x $0.5^(2)$ = 3.142 m^2

terminal velocity is gotten from the relationship

$Vt = \sqrt{(2mg)/(pACd) }$

where g = acceleration due to gravity = 9.81 m/s^2

imputing values into the equation

$Vt = \sqrt{(2*10*9.81)/(1.22*3.142*0.8) }$ = 7.99 m/s

If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3

$Vt = \sqrt{(2mg)/(pACd) }$

$Vt = \sqrt{(2*10*9.81)/(0.736*3.142*0.8) }$ = 10.298 m/s

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Learn more about the Coefficient of kinetic friction here:

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Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.

Answers

Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

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In a game of tug of war, a rope is pulled by a force of 182 N to the right and by a force of 108 N to the left. Calculate the magnitude and direction of the net horizontal force on the rope.