Acceleration is mathematically defined as; ΔV/t.
<h3>What is acceleration?</h3>
Acceleration is defined as the ratio of the change in velocity to time. It is defined as the extent to which the velocity changes within a given time interval.
The question is incomplete so we can not arrive at the final answer. However, acceleration is mathematically defined as; ΔV/t.
Learn more about acceleration: brainly.com/question/2437624
Answer:
1.) Micrometres screw gauge
2.) Tape rule.
Explanation:
Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.
what are the best instruments to use ?
To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.
And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.
Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
![V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]](https://tex.z-dn.net/?f=V%20%3D%200.15%2A0.002%2A0.1%5C%5CV%20%3D%200.00003%20%5Bm%5E%7B3%7D%20%5D)
And the mass m = 4 [gramm] = 0.004 [kg]
![Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]](https://tex.z-dn.net/?f=Ro%20%3D%200.004%2F0.00003%5C%5CRo%20%3D%20133%20%5Bkg%2Fm%5E%7B3%7D%5D)
Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules