Answer: the hunter should aim his rifle 3,422 meters above the deer to hit it.
Explanation: We can solve this problem using projectile motion equations. Let's first identify the known variables:
Distance to the deer along the line of sight (horizontal distance): d = 10,181 m
Height of the hill: h = 90 m
Muzzle velocity of the gun: v0 = 100 m/s
Acceleration due to gravity: g = 10 m/s^2
We want to find the vertical distance (height) between the hunter's location and the deer, which we can call y. We can use the following equation for the vertical motion of a projectile:
y = v0tsin(theta) - (1/2)gt^2
where t is the time it takes for the bullet to reach the deer, and theta is the angle between the line of sight and the horizontal (the angle at which the rifle is aimed).
To find t, we can use the fact that the time of flight of a projectile is given by:
T = 2v0sin(theta)/g
We can solve for sine(theta) using the fact that the horizontal distance is related to the time of flight by:
d = v0*cos(theta)*T
Combining these equations, we get:
sin(theta) = d/(2Tv0)
cos(theta) = sqrt(1 - sin^2(theta))
t = T/2
Substituting these values back into the equation for y, we get:
y = (d*t/tan(theta)) - (1/2)gt^2
Now we just need to plug in the numbers and solve for y:
sin(theta) = 90/10,181
theta = arcsin(sin(theta)) = 0.523 degrees
cos(theta) = cos(0.523) = 0.999
T = 2100sin(theta)/g = 3.189 seconds
t = T/2 = 1.595 seconds
y = (10,181*t/tan(theta)) - (1/2)gt^2 = 3,422 meters
Therefore, the hunter should aim his rifle 3,422 meters above the deer to hit it. Note that this assumes no air resistance and that the rifle is perfectly aimed. In reality, the bullet would be affected by air resistance and the trajectory would depend on the accuracy of the rifle and the shooter's skill.
Excess charge of 9. 0 is placed 0. 17 m away with charge of -3. 2
By calculating this expression [tex]8.99 * 10^9 N m^2/C^2[/tex] and using the given values, we can find the force between the charges.
When an excess charge of 9.0 (let's call it Q1) is placed 0.17 meters away from a charge of -3.2 (Q2), we can determine the force between them using Coulomb's Law.
Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
The formula for the electric force (F) between two charges is given by:
[tex]F = k * |Q1 * Q2| / r^2[/tex]
where k is the electrostatic constant.
Substituting the given values into the formula, we have:
[tex]F = k * |9.0 * (-3.2)| / (0.17)^2[/tex]
To determine the exact value of the force, we need the value of k. The electrostatic constant, k, is approximately [tex]8.99 * 10^9 N m^2/C^2[/tex].
Therefore, by calculating the expression using the given values, we can find the force between the charges.
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an object is dropped from a cliff. after 3 seconds, its acceleration is _____ m/s 2
According to the question, The acceleration of an object in free fall near the surface of the Earth is approximately 9.8 m/s².
When an object is dropped from a cliff, it experiences gravitational acceleration directed towards the Earth. This acceleration is constant and equal to the acceleration due to gravity, which is approximately 9.8 m/s² near the surface of the Earth. Therefore, after 3 seconds, the acceleration of the object will still be approximately 9.8 m/s². The acceleration remains constant unless other forces, such as air resistance, come into play. This is because the acceleration due to gravity remains constant unless other forces, such as air resistance, significantly affect the motion.
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what is the decay rate of americium in gs-1? that's inverse gigaseconds.
The decay rate of Americium-241 in inverse gigaseconds is approximately 7.309 x 10⁻¹⁸ Gs⁻¹.
The decay rate of Americium is typically expressed in terms of its half-life, which is the time it takes for half of the radioactive substance to decay. The half-life of Americium-241, for example, is approximately 432.2 years.
To determine the decay rate in inverse gigaseconds (Gs⁻¹), we need to convert the half-life from years to gigaseconds.
1 year is equivalent to 3.16888 x 10⁻¹⁷ gigaseconds (Gs). Therefore, the half-life of Americium-241 is approximately 1.3684 x 10¹⁶ Gs.
The decay rate is the reciprocal of the half-life, so the decay rate of Americium-241 in inverse gigaseconds (Gs⁻¹) would be approximately 7.309 x 10⁻¹⁸ Gs⁻¹.
Therefore, the decay rate of Americium-241 in inverse gigaseconds is approximately 7.309 x 10⁻¹⁸ Gs⁻¹.
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