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Firlakuza [10]
3 years ago
7

Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st

raight up, ball 2 is fired straight down, and ball 3 is fired horizontally. Rank in order from largest to smallest the speeds of the balls, v1, v2, and v3, just before each ball hits the ground. a. v1 > v2 > v3 b. v3 > v2 > v1 c. v2 > v3 > v1 d. v1 = v2 = v3
Physics
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

d) v1 = v2 = v3

Explanation:

This can be answered using conservation of energy. We calculate the mechanical energy E=K+U (sum of kinetic and gravitational potential energies) at the original and final points, and impose they are equal.

At the original point we have, for the three balls:

E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh

At the final point we have, for the three balls:

E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}

Since we have E_i=E_f, and E_i is the same for all balls, then E_f is the same for all balls, which means that v_f, the final velocity, is the same for all balls.

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Which of the following describes the charge of an atom before any electrons are transferred?
gregori [183]

The answer is neutral charge. An atom element will always and has to be stable, in order for this state to happen. The charge of an electron has to be neutral. For atom with neutral charge, the proton will always equal the number of electron.

8 0
3 years ago
According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
Bond [772]

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
3 years ago
You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes. That is half of your blood
Zina [86]

Answer:

8 times

Explanation:

Given that You have about 10 quarts of blood in your body. At REST your heart pumps about 5 quarts each minutes.

That means the heart will pump 10 quarts in 2 minutes.

That is half of your blood volume per minute.

If during exercise it can pump 40 quarts per minute, that is, 80 quarts in 2 minutes.

To know how many times does all of your blood complete the cycle around your body during exercise, you must divide 80 quarts by 10 quarts. That is,

80 / 10 = 8

Therefore, your blood complete the cycle around your body 8 times during the exercise.

3 0
3 years ago
Beaker A contains 100 mL of water at a temperature of 25 °C. Beaker B contains 100 mL of water at a temperature of 60 °C. Which
Novosadov [1.4K]

Answer:

Only option A is correct. Beaker A has lower kinetic energy than beaker B.

Explanation:

Step 1: Data given

Beaker 1 has a volume of 100 mL at 25 °C

Beaker B has a volume of 100 mL at 60 °C

Thermal energy = m*c*T

Thermal energy beaker A = 100 grams*4.184 * 25°C

Thermal energy beaker B = 100 grams *4.184*60°C

⇒ Since both beakers contain the same amount of water, the thermal energy depends on the temperature.

Since beaker B has a higher temperature, it has a higher thermal energy than beaker A

When we heat a substance, its temperature rises and causes an increase in the kinetic energy of its constituent molecules. Temperature is, in fact, a measure of the kinetic energy of molecules.

This means beaker B has a higher kinetic energy than beaker A

Potential energy doesn't depend on temperature. this means the potential energy of beaker A and beaker B is the same.

a. Beaker A has lower kinetic energy than beaker B. This is correct.

b. Beaker A has higher thermal energy than beaker B. This is false.

c. Beaker A has higher potential energy than beaker B. This is false.

d. Beaker A has lower potential energy than beaker B. This is false

e. Beaker A has higher kinetic energy than beaker B. This is false.

3 0
2 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
2 years ago
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