It changes the motion as Newton's second law of motion states that a force, acting on an object, will change its velocity by changing either its speed or its direction or both. If your basketball goes rolling into the street and is hit by a bike, either the ball will change direction or its speed or both.
Density = 2.7g/cm3
mass =8.1grams
from density =(mass)/volume
we can determine the volume of aluminium by simply changing the subject of the subject of the formula from density to volume
so we have
volume=mass/density
=(8.1grams)/(2.7g/cm3)
=3cm3
When an astronaut brings a cube from the Earth to the Moon, the inertial mass remains constant, but the weight decreases.
<h3>What is the difference between mass and weight?</h3>
Mass of the body is defined as the amount of matter a body have. It is denoted by m and its unit is kg.
Weight is defined as the amount of force an object expert on the surface. It is given as the product of mass and the gravitational pull.
When an astronaut brings a cube from the Earth to the Moon, the inertial mass remains constant, but the weight decreases.Because the value of the gravitational acceleration is different on the moon.
Hence, option D is correct.
To learn more about the mass, refer to the link;
brainly.com/question/19694949
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Answer:
The force will be "5,488". A further solution is provided below.
Explanation:
The given values are:
speed,
v = 100m/s
mass,
m = 80 kg
acceleration,
a = 7
Now,
Radius will be:
⇒ 
⇒ 
⇒ 
⇒ 
Force will be:
⇒ 
⇒ 
⇒ 
Answer:
a) 7200 ft/s²
b) 140 ft
c) 3.7 s
Explanation:
(a) Average acceleration is the change in velocity over change in time.
a_avg = Δv / Δt
We need to find what velocity the puck reached after it was hit by the hockey player.
We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:
v² = v₀² + 2a(x − x₀)
(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)
v₀² = 5200 ft²/s²
v₀ = 20√13 ft/s
So the average acceleration impacted to the puck as it is struck is:
a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)
a_avg = 2000√13 ft/s²
a_avg ≈ 7200 ft/s²
(b) The distance the puck travels before stopping is:
v² = v₀² + 2a(x − x₀)
(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)
x = 140 ft
(c) The time the puck takes to travel 10 ft without friction is:
t = (10 ft) / (20√13 ft/s)
t = (√13)/26 s
The time the puck travels over the rough ice is:
v = at + v₀
(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)
t = √13 s
So the total time is:
t = (√13)/26 s + √13 s
t = (27√13)/26 s
t ≈ 3.7 s
