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dalvyx [7]
4 years ago
6

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

angle of projection are kept the same. On Earth, the javelin covers a distance y. What distance would it cover on the moon? (The moon’s gravity is about 1/6th that of Earth.) A) 6y B) 6/y C) y/6 D) (y + 6)/2
Physics
2 answers:
strojnjashka [21]4 years ago
7 0
The answer would be A: 6y
Great explanation below.  Keep studying ole Chap
SSSSS [86.1K]4 years ago
6 0
Since the gravity at the moon is around 6 times weaker, we have that the downwards pull is around 6 times smaller. In equations, we have that g of the moon=g of Earth/6 and since the weight is F=m*g, the force is 1/6th of the force on earth. Hence, the distance is going to be around 6 times larger than the distance traversed on Earth and thus the correct relation is A=6y.
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Caffeine concentration is 1.99 mg/oz how many cans would be leathal if 10g was leathal and there where 12oz in a can
katovenus [111]

The number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

<h3>How to convert mass?</h3>

According to this question, caffeine concentration is 1.99 mg/oz.

1.99 milligrams can be converted to grams as follows:

1.99milligrams ÷ 1000 = 0.00199grams

This means that 0.00199grams per oz is the caffeine concentration.

If there were 12 oz in a can, then, 0.00199grams × 12 = 0.02388 grams in 1 can.

This means that if 10grams is considered lethal, 10grams ÷ 0.02388 grams = 419 cans would be lethal for consumption.

Therefore, the number of cans that would be considered lethal if 10g was lethal and there where 12oz in a can is 419 cans.

Learn more about conversion factor at: brainly.com/question/14479308

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5 0
2 years ago
Under what conditions will an object experience a gravitational force?
Radda [10]
A) an object with mass > 0 in a gravitational field
b) an object with an electric charge not 0 in an electric field
c) a moving object with an electric charge not 0 in a magnetic field
6 0
3 years ago
A 2.5 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnt
emmainna [20.7K]

The concepts required to solve this problem are those related to the conservation of the angular momentum and the moment of inertia of the disk. We will begin by calculating the moment of inertia of the disc, then the moment of inertia of the disc after the two two blocks hits and sticks to the edges of the turn table. In the end we will apply the conservation theorem.

The radius is given as,

R = \frac{20cm}{2} = 10cm = 0.1m

When a block falls from above and sticks to the turn table, the moment inertia of the turntable increases.

Since two blocks are stick to the turn table, the total final moment of inertia of the turntable is the sum moment of inertias of individual turntable, and two blocks.

I_1 = \frac{1}{2} MR^2

I_1 = \frac{1}{2} (2.5)(0.1)^2

I_1 = 0.0125kg \cdot m^2

The moment of inertia of each block is

I_0 = mR^2

Total moment of inertia of two block is

I_0' = 2mR^2

The final moment of inertia of the turn table is

I_2 = I_1 +I'0

I_2 = I_1 +2mR^2

I_2 = 0.01kg\cdot m^2 + 2(500*10^{-3}kg)(0.1m)^2

I_2 = 0.0225kg\cdot m^2

From the conservation of the angular momentum, the initial angular momentum of the system is equal to final angular momentum of the system,

Rearrange the equation we have that

I_1\omega_2 = I_2\omega_2

\omega_2 = \frac{I_1\omega_2}{I_2}

\omega_2 = \frac{0.01*150rpm}{0.0225}

\omega_2 = 66.67rpm

The magnitude of the turntable's angular velocity is 66.67rpm

3 0
3 years ago
What is the general relationship between the size of an atom and its first ionization energy?
just olya [345]
The smaller the atom,the larger the first I.E.
3 0
3 years ago
De Vico Comet orbits the Sun every 74.0 years and has an orbital eccentricity of 0.96. Find the comet's average distance from th
Debora [2.8K]

Answer: The comet's average distance from the sun is 17.6AU

Explanation:

From Kepler's 3rd Law, P^2=a^3

Where P is period in years

and a is length of semi-major axis or the average distance of the comet to the sun.

Given the orbital period to be 74 years

74^2 =a^3

5476 = a^3

Cube root of 5476 =a

17.626 = a

Approximately a= 17.6 AU

5 0
4 years ago
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