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Alex throws a 0.15-kg rubber ball down onto the floor. The ball’s speed just before impact is 6.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for 0.025 s, what is the magnitude of the average force applied by the floor on the ball?

Answers

Answer 1

Answer:

The magnitude of the average force applied by the floor on the ball is 60 N

From the question given above, the following data were obtained:

Mass (m) = 0.15 Kg

Initial velocity (u) = 6.5 m/s

Final velocity (v) = 3.5 m/s

Time (t) = 0.025 s

Force (F) =?

The magnitude of the averageforce applied by the floor on the ball can be obtained as follow:

$F = (m(v + u) )/(t) \n\nF = (0.15(3.5 + 6.5) )/(0.025) \n\nF = (0.15(10))/(0.025)\n\n$

F = 60 N

Thus, the magnitude of the average force applied by the floor on the ball is 60 N

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Answer 2

Answer:

Answer: Change in ball's momentum is 1.5 kg-m/s.

Explanation: It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

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If the rise and fall of your lungs is considered to be simple harmonic motion, how would you relate the period of the motion to your breathing rate (breaths per minute)? Breaths per minute is an angular frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is its reciprocal. Breaths per minute is an angular frequency. The period is its reciprocal.

Answers

Answer:

Breaths per minute is a frequency. The period is its reciprocal.

Explanation:

In simple harmonic motion, a period (T) is the time taken for one point to start in a position and reach that position again, in other words to complete a cycle or lapse. In this case, a period is the time one takes from starting to inspire the air to releasing all of it from the lungs.

In simple harmonic motion, the frequency (f) is how many times a point completes a cycle or lapse in one unity of time (could be one second, one minute, one hour, etc). In this case, the frequency is how many times one breathes in one minute. This is the breathing rate, since it is breathings per minute. Breaths per minute is a frequency.

Period (T) and frequency (f) relate to each other in the following formulae: $T=(1)/(f)$ or $f=(1)/(T)$ .

Therefore, breaths per minute is a frequency, and since it is related to the period, we say the period is reciprocal to it.

The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)

Answers

This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

$W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)$

Taking the derivative with respect to actual temperature, we get:

$(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n$

where,

dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h

Therefore,

$dW=(-1)(0.6215-0.3965(18)^(0.16))$

dW = - 1.25°C

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Answer:

Δw=1.25°C

Explanation:

Given that

$w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)$

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

$(dw)/(dT)=0.6215 +0.3965v^(0.16)$

$(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)$

Now by putting the values v=19 km/hr and ΔT=1

$(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)$

Δw=1.25°C

So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

A stock person at the local grocery store has a job consisting of the following five segments:1) picking up boxes of tomatoes from the stockroom floor

2)accelerating to a comfortable speed.

3) Carring the boxes to the tomato display at constant speed.

4)decelerating to a stop.

5) lowering the boxes slowly to the floor.

During which of the five segments of the job does the stock person do positive work on the boxes?

A) (2) and (3)

B(1) and (2)

C) (1) only

D) (1), (2), (4) and (5)

E) (1) and (5)

Answers

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:

Answers

Explanation:

Charge on proton, q₁ = e

Charge on alpha particles, q₂ = 2e

The magnetic force is given by :

$F=qvB\ sin\theta$

Here, $\theta=90=sin(90) = 1$

For proton, $F_p=ev_pB$ ..........(1)

For alpha particle, $F_a=2ev_aB$ ..........(2)

Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,

$ev_pB=2ev_aB$

$(v_p)/(v_a)=(2)/(1)$

So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.

Final answer:

If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.

Explanation:

The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.

The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.

This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.

Therefore, the proton is moving twice as fast as the alpha particle.

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A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying

Answers

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

p = q s

s = p / q

let's calculate

s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

s = 0.625 10⁻²¹ m

s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

Beginning 156 miles directly east of the city of Uniontown, a truck travels due south. If the truck is travelling at a speed of 31 miles per hour, determine the rate of change of the distance between Uniontown and the truck when the truck has been travelling for 81 miles. (Do not include units in your answer, and round to the nearest tenth.)