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3 years ago

Jamal is at Six Flags playing at the arcade. At one booth he throws a 0.50 kg ball

1 answer:

8 0

were solving for v velocity of the ball after it has hit the bottle. a. momentum ->p=mv->ball + bottle momentum during hit = ball + bottle momentum after hit-> ball (.5*21)+ bottle (.2*0) (it's 0 because the the bottle is standing still) = ball after hit (.5*v)+bottle after hit (.2*30) -> 10.5+0=.5v+6 ->4.5=.5v->v=9m/s

b. if bottle was heavier the ball would be slower so final velocity would decrease

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Which of the following is an example of a technological solution to water pollution in a local pond?

dexar [7]

Sure: ths is called protyping and lets yu get a sende fo the effeciveness tof the cahnge and the cost of the change.

<span>A - just like the scientific principle syu what to know what other know or have learned. Example would it be silly to build a nuclear power de-salinatiztion plant when a dam in the mountains wuld dothe savme thng and perhaps have the advatage of using local labor and preveinting floods and givng of hydro eletic power. </span>

<span>A a process is a technological soultion it uses tools, machines, chemical to effect an otu come.</span>

8 0

3 years ago

Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis. The magnitude of q1 is 3 times the magnitu

Alex777 [14]

d = distance between the two point charges = 60 cm = 0.60 m

r = distance of the location of point "a" where the electric field is zero from charge $q_{1}$ between the two charges.

$q_{1}$ = magnitude of charge on one charge

$q_{2}$ = magnitude of charge on other charge

$q_{1}$ = 3 $q_{2}$

$E_{1}$ = Electric field by charge $q_{1}$ at point "a"

$E_{2}$ = Electric field by charge $q_{2}$ at point "a"

Electric field by charge $q_{1}$ at point "a" is given as

$E_{1}$ = k $q_{1}$ /r²

Electric field by charge $q_{2}$ at point "a" is given as

$E_{2}$ = k $q_{2}$ /(d-r)²

For the electric field to be zero at point "a"

$E_{2}$ = $E_{1}$

k $q_{2}$ /(d-r)² = k $q_{1}$ /r²

$q_{2}$ /(d-r)² = 3 $q_{2}$ /r²

1/(0.60 - r)² = 3 /r²

r = 0.38 m

r = 38 cm

8 0

3 years ago

Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0

3 years ago

The Curiosity rover now on Mars analyzed rocks and found magnesium to have the following isotopic composition.

emmasim [6.3K]

Answer:

A. number of neutrons of Magnesium Mg = 13

B. The average mass of Mg = 22.29 amu

C. the magnesium composition on Mars is not the same as that on Earth.

Explanation:

Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.

The atomic number of Magnesium Mg = 12

The atomic number of an element is the number of protons present in the atomic nucleus of the element

i.e Atomic number = number of protons = 12

The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.

Mass number = number of protons + number of neutrons

Given that the mass number of Mg = 25

Then;

25 = 12 + number of neutrons

25 - 12 = number of neutrons

13 = number of neutrons

number of neutrons of Magnesium Mg = 13

B. What is the average atomic mass of magnesium in these rocks?

The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Therefore the average atomic mass of magnesium can be calculated as:

= $\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}$

= $\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}$

= $\mathtt{\dfrac{2229.17774 }{100}}$

The average mass of Mg = 22.29 amu

C. Is the magnesium composition on Mars the same as that on Earth? Explain.

The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.

There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.

7 0

3 years ago

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo

S_A_V [24]

Answer:

A. Z = 185.87Ω

B. I = 0.16A

C. V = 1mV

D. VL = 68.8V

E. Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula: