Answer:
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How is Mass or volume change affect density?: Density is an intensive property of the material or substance and depends upon the relationship between the mass and volume. Unless the mass changes in relation to the volume, the density will not change.
Are mass and volume related?: Mass and volume are two units used to measure objects. Mass is the amount of matter an object contains, while volume is how much space it takes up.We can say that the volume of the object is directly proportional to its mass. As the volume increases the mass of the object increases in direct proportion.
How can density of an object be determined?: If the mass of an object increases then its density increases because density is directly proportional to mass.
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Explanation:
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When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.
Projectile is a missile designed to be fired from a rocket or gun.
A projectile is the object that is propelled by the application of an external force and then moves freely under the influence of gravity and air resistance.
The range is defined as the distance between the launch point and the point where the projectile hits the ground.
The height from the ground at the top most position of projectile is referred to as maximum height.
When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.
Learn more about maximum height click here brainly.com/question/6261898
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Answer:
46.22 cm
Explanation:
The focal refraction, fr is given by
The focal red light is given by
![fv = \frac {c}{(1.605 - 1)} = \frac {c}{0.605}](https://tex.z-dn.net/?f=fv%20%3D%20%5Cfrac%20%7Bc%7D%7B%281.605%20-%201%29%7D%20%3D%20%5Cfrac%20%7Bc%7D%7B0.605%7D)
![\frac {fv}{fr} = \frac {0.572}{0 .605} = 0.945455](https://tex.z-dn.net/?f=%5Cfrac%20%7Bfv%7D%7Bfr%7D%20%3D%20%5Cfrac%20%7B0.572%7D%7B0%20.605%7D%20%3D%200.945455)
and making fr the subject we obtain
![fr = \frac {image * object}{(image + object)} = \frac {24.00 * 55} {(24.0 + 55)} = 16.70886 cm](https://tex.z-dn.net/?f=fr%20%3D%20%5Cfrac%20%7Bimage%20%2A%20object%7D%7B%28image%20%2B%20object%29%7D%20%3D%20%5Cfrac%20%7B24.00%20%2A%2055%7D%20%7B%2824.0%20%2B%2055%29%7D%20%3D%2016.70886%20cm)
fv = 0.945455* 16.70886 cm = 15.79747 cm
![image = \frac {object * f} {(object - f)} = \frac {15.79747 * 24.0}{(24.0 - 15.79747)} = 46.22222 cm](https://tex.z-dn.net/?f=image%20%3D%20%5Cfrac%20%7Bobject%20%2A%20f%7D%20%7B%28object%20-%20f%29%7D%20%3D%20%5Cfrac%20%7B15.79747%20%2A%2024.0%7D%7B%2824.0%20-%2015.79747%29%7D%20%3D%2046.22222%20cm)
Therefore, violet image is approximately 46.22 cm
Answer:
a) J = F t = 40 * .05 = 2 N-s
b) J = 2 N-s momentum changed by 2 N-s
c) Initial momentum appears to be zero
J = change in momentum = m v2 - m v1 = m v2 = 2 N-s
v2 = J / m = 2 / .057 = 35 m/s
d) if the impulse time was increased and the average force remained the same then the change in momentum would increase with a corresponding increase in velocity attained - note the increase in v2 in part c)
It’s very big and very small numbers