PHYSICS COLLEGE

After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.0 m. Determine the magnitude of the impulse delivered to the ball by the floor.

Answers

Answer 1

Answer:

Given Information:

Mass of ball = m = 0.320 kg

Initial height = h₁ = 19 m

Final height = h₂ = 15 m

Required Information:

Impulse = I = ?

Answer:

Impulse = 11.77 kg.m/s

Explanation:l

We know that impulse is equal to change in momentum

I = Δp

I = p₁ - p₂

I = mv₁ - mv₂

I = m(v₁ - v₂)

Where m is the mass of ball, v₂ is the final velocity of the ball, and v₁ is the initial velocity of the ball.

So first we need to find the initial and final velocities of the ball

The relation between initial potential energy and final kinetic energy before the collision is given by

PE₁ = KE₂

mgh₁ = ½mv₂²

gh₁ = ½v₂²

v₂² = 2gh₁

v₂ = √2gh₁

v₂ = √2*9.8*19

v₂ = 19.3 m/s

The relation between initial kinetic energy and final potential energy after the collision is given by

KE₁ = PE₂

½mv₁² = mgh₂

½v₁² = gh₂

v₁² = 2gh₂

v₁ = √2gh₂

v₁ =√2*9.8*15

v₁ = 17.15 m/s

Finally, we can now find the magnitude of the impulse delivered to the ball by the floor.

I = 0.320(17.5 - (-19.3))

I = 11.77 kg.m/s

Answer 2

Answer:

Answer:

$Imp = 11.666\,(kg\cdot m)/(s)$

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

$(0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (19\,m) = (1)/(2)\cdot (0.320\,kg)\cdot v_(A)^(2)$

$v_(A) \approx 19.304\,(m)/(s)$

After collision:

$(1)/(2)\cdot (0.320\,kg)\cdot v_(B)^(2) = (0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (15\,m)$

$v_(B) \approx 17.153\,(m)/(s)$

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

$Imp = (0.32\,kg)\cdot [(17.153\,(m)/(s) )-(-19.304\,(m)/(s) )]$

$Imp = 11.666\,(kg\cdot m)/(s)$

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Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

Answers

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, with a standard deviation of σ = 0.5°C. How many readings must be taken so that the probability is 0.90 that the average is within ±0.1◦C of the actual temperature? Round the answer to the next largest whole number.

Answers

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

$0.9 = P(|\bar{x}-\mu|<0.1)$

$0.9 = P(\frac{|\bar{x}-\mu|}{(\sigma)/(√(n))}<(0.1)/((\sigma)/(√(n))))$

$0.9 = P(|Z|<(0.1)/((\sigma)/(√(n))))$

For a confidence level of 90% our $Z_(critic)$ is 1.645

Therefore,

$(0.1)/((\sigma)/(√(n))) = 1.645$

Substituting for $\sigma = 5$ and re-arrange for n, we have that n is equal to

$n=((1.645\sigma)/(0.1))^2$

$n=((1.645)^2(0.5)^2)/(0.1^2)$

$n=67.65$

$n=68$

We need to make 68 readings for have a probability of 90% and our average is within $0.1\°\frac$

A substance that does NOT conduct an electric current when it forms a solution is a(n) ____. a electrolyte

b nonelectrolyte

c liquid

d solid

Answers

Answer:

B. Nonelectrolyte.

Explanation:

Nonelectrolytes do not dissociate into ions in solution, hence, nonelectrolyte solutions don't conduct electricity.

A non-electrolyte doesn’t conduct electric current even when it forms a solution.

Answer: Option B

Explanation:

Where electrolytes are defined as the compounds that can conduct electric current with mobile ions existing in its solution, non-electrolytes are the compounds that don’t behave the same either in the aqueous solution or in the molten state.

This is all because these compounds don’t produce mobile ions to flow from one electrode to the other and hence conduct electric flow in the solution. Sugar and ethanol are the best examples of non-electrolytes that don’t induce electric current even after getting dissolved in water.

Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?

Answers

Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

v² = u² + 2as

v² = 44.50² + 2 x (-9.81) x 71.32

v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

v = u + at

24.10 = 44.50 - 9.81 t , t = 2.07 s

-24.10 = 44.50 - 9.81 t , t = 6.99 s

Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

s = ut + 0.5 at²

71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answers

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\n\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=(1)/(3)mL^2$

substituting known values:

$I=(1)/(3) (0.9kg)(0.95m)^2\nI=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=(\omega)/(t)$

susbtituting known values:

$\alpha=(16.336rev/s)/(0.2s) \n\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\n\tau=22.13Nm$

Final answer:

To calculate the torque, we need to use the formula: Torque = Moment of Inertia * Angular Acceleration. By approximating the bat as a uniform rod and using its length and mass, we can find the moment of inertia. Then, using the given angular velocity, we can calculate the angular acceleration. Finally, we can determine the torque by multiplying the moment of inertia by the angular acceleration.

Explanation:

To compute the torque the player applies to one end of the bat, we need to use the formula:

Torque = Moment of Inertia * Angular Acceleration

Given that the bat is approximated as a uniform rod and we know its length and mass, we can calculate the moment of inertia. Then, using the given angular velocity, we can compute the angular acceleration. Finally, we can find the torque by multiplying the moment of inertia by the angular acceleration.

Learn more about Torque here:

brainly.com/question/33222069

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why did thomson's from experermenting with cathode rays cause a big change in scientific thought about atoms

Answers

Answer:

His results gave the first evidence that atoms were made up of smaller particles.

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