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Jobisdone [24]
3 years ago
7

A rubber ball of mass 18.5 g is dropped from a height of 1.90 m onto a floor. The velocity of the ball is reversed by the collis

ion with the floor, and the ball rebounds to a height of 1.45 m. What impulse was applied to the ball during the collision?
Physics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

J=0.211kg.m/s

Explanation:

The impulse-momentum theorem states:

J=\Delta p\\J=m(v_a-v-b)\\where:\\m=mass\\v_a=velocity\_after\\v_b=velocity\_before

The velocity before the impact is given by:

(v_b)^2=2.a.\Delta y\\v_b=\sqrt{2*(9.8m/s^2)1.90m}=6.10m/s(-\hat{j})

For the velocity after the impact:

(v_f)^2=(v_a)^2+2.a.\Delta y\\(0)^2=(v_a)^2+2.(-9.8m/s^s).(1.45m)\\\\v_a=\sqrt{2*9.8m/s^2*1.45m}\\v_a=5.33m/s(\hat{j})

so:

J=18.5*10^{-3}kg(5.33m/s(\hat{j})-6.10m/s(-\hat{j}))\\\\J=18.5*10^{-3}kg(5.33m/s(\hat{j})+6.10m/s(\hat{j}))\\\\J=0.211kg.m/s

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A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a
Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged  (x)= 0.13 m

now,

weight of the mass attached = Kx

             m g = k x

             0.35 x 9.8 = k x 0.13

                k = 26.38 N/m

now, using conservation of energy

 \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2

 v = \sqrt{\dfrac{kx'^2}{m}}

 v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}

 v = \sqrt{1.6958}

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6 0
3 years ago
In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

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8 0
3 years ago
Which properties of plastics allow them to be solutions to many complex problems in the world? Check all that apply. A.chemicall
grigory [225]

the answers are A, B, C, E  HOPED THIS HELPED

3 0
3 years ago
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The electrons lost from chlorophyll photooxidation are replaced by the oxidation of water. how many electrons are generated from
White raven [17]

There are 2 electrons generated from the oxidation of one water molecule.

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1 year ago
Tina has been dieting for a total of 13 weeks she lost 3 lb on the first week of her diet but gain the back of pound on the seco
PolarNik [594]

Answer:

D. 24 lb

Explanation:

Tina has been dieting for 13 weeks

First week she lost 3 pounds

Next week she gained 1 pound and did not lose any. This will be subtracted as she has gained a pound

The remaining 11 weeks she lost 2 pounds per week

Weight lost in the 11 weeks = 11×2 =22 pounds

Total weight lost

3-1+22 = 24 lb

Tina has lost 24 pounds in total during the 13 weeks

6 0
3 years ago
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