Answer:
Systematic errors.
Explanation:
The density of the aluminium was calculated by a human and this is not natural but can be due to errors in the calibration of the scale for measuring the weight or taking readings from the measuring cylinder.
Random errors are natural errors. Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment. Systematic errors are due to imprecision or problems with instruments.
Answer:
Explanation:
= Cambio en la longitud de la cuerda = 0.25 cm
T = tensión en cuerda
A = Área de la cadena = 
d = Diámetro de la cuerda = 0.2 cm
L = Longitud original de la cuerda = 1.6 m
El cambio de longitud de una cuerda viene dado por
La tensión en la cuerda es 
.
The force result in stretching the spring 10.0 centimeters is 2.5N.
<h3>
What is Hooke's law?</h3>
If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
Learn more about hooke's law.
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By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
Answer:
All fraction of kinectic energy is lost to barrel of a spring gun of mass 1.8 kg
Explanation:
A ball of mass 0.50 kg is fired with velocity 160 m/s ...
The kinetic energy is given by 1/2mv²
Kinectic energy of the ball = 1/2 *0.5*160²
Kinectic energy = 1/4 *25600
Kinectic energy = 6400 joules.
If no energy is lost to fiction, and the ball sticks to a barrel of a spring gun of mass 1.8 kg with initial velocity zero, all kinetic energy is lost to the barrel of a spring gun of mass 1.8 kg.
