If the temperature of an iron sphere is increased A)its density will decrease .B)its volume will decrease. C)its mass will decre
ase. D)its density will increase. E)its density will remain unchanged
2 answers:
Answer:
A) Its density will decrease
Explanation:
When an object is heated, its volume increases. This is due to the fact that the particles in the medium vibrate more (if it is a solid) or they move more (if it is a liquid or a gas), therefore they tend to occupy a larger space.
At the same time, the mass of the object does not change, because the mass just represents the amount of matter contained in the object, so it does not increase/decrease at different temperatures.
The density of an object is defined as the ratio between the mass (m) and the volume (V):
We said that the mass remains unchanged while the volume increases: since the density is inversely proportional to the volume, this means that the density decreases.
If the temperature of an iron sphere is increased , then A) its density will decrease
<h3>Further explanation</h3>This problem is about Density.
Density is the ratio of mass to the volume of the object.
<em>ρ = density of object ( kg / m³ )</em>
<em>m = mass of object ( kg )</em>
<em>V = volume of object ( m³ )</em>
When the iron sphere is heated, it will expand. This expansion causes an increase in the volume of the sphere.
The volume expansion of the iron sphere could be calculated using this following formula.
where:
<em>V = final volume after expansion ( m³ )</em>
<em>Vo = initial volume before expansion ( m³ )</em>
<em>α = coefficient of expansion ( /K )</em>
<em>Δt = change in temperature ( K )</em>
The mass of the iron ball does not change, so according to the formula above, the density of the iron sphere will decrease.
This is because the density is inversely proportional to the volume of the object.
The greater the volume of the object, the smaller the density of the object.
<h3>Learn more</h3>- Velocity of Runner : brainly.com/question/3813437
- Kinetic Energy : brainly.com/question/692781
- Acceleration : brainly.com/question/2283922
- The Speed of Car : brainly.com/question/568302
Grade: High School
Subject: Mathematics
Chapter: Density
Keywords: Temperature , Density , Iron , Sphere , Volume , Mass
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Answer:
a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)
b. Apparent weight at the top is approximately 519.06 N
Apparent weight at the bottom is approximately 558.94 N
Explanation:
a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly
b. The given parameters are
The radius of the Ferris wheel, r = 7.2 m
The period for one complete revolution, t = 28 seconds
The angle covered in one revolution, θ = 2·π radian
The mass of the person riding on the Ferris wheel, the passenger = 55 kg
Therefore, we have;
The angular speed, ω = Δθ/Δt = 2·π/(28)
From which we have;
Centripetal force, = m × ω² × r
Substituting the known values, we have = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N
The centripetal force, = 19.94 N always acting outward from the center
Weight = Mass × Acceleration due to gravity
The weight of the passenger = 55 kg × 9.8 m/s² = 539 N
The weight of the passenger = 539 N always acting downwards
At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards
Therefore;
The net force, which is the apparent weight of the passenger at the top = 539 N - 19.94 N ≈ 519.06 N
Apparent weight at the top ≈ 519.06 N
At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards
Therefore;
The net force at the bottom, which is the apparent weight of the passenger at the bottom = 539 N + 19.94 N ≈ 558.94 N
Apparent weight at the bottom ≈ 558.94 N.
Answer:
The shearing stress is 10208.3333 Pa
The shearing strain is 0.25
The shear modulus is 40833.3332 Pa
Explanation:
Given:
Block of gelatin of 120 mm x 120 mm by 40 mm
F = force = 49 N
Displacement = 10 mm
Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?
The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:
A = 40*120 = 4800 mm² = 0.0048 m²
The shearing stress:
The shearing strain is defined as the tangent of the displacement that the block over its length:
Finally, the shear modulus is the division of the shearing stress over the shearing strain: