Answer:
<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/
)
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/
F = 122.63 N
<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>
<span>Examples of Newton's 2nd Law If you
use the same force to push a truck and push a car, the car will have
more acceleration than the truck, because the car has less mass. It is
easier to push an empty shopping cart than a full one, because the full
shopping cart has more mass than the empty one.</span>
from slide share.com
Answer:
118kg
Explanation:
answered
Given
density of the cube= 8050 kg/m3 .
length of the sizes of the cube=24.5 cm
We can convert the length to cm for unit consistency.
It's length =24.5 cm =0.245m
✓ the length of sizes of the cube is the same, then the volume can be calculated as
Volume= L^3
= (0.245m)^3
=0.01470625 m^3
✓ but we know that
Density = mass/ volume
Then,
Mass= (Volume × density)
= (0.01470625)(8050)
= 118 kg
Hence, the mass of the cube is 118 kg
Is an imbalance of electric charges within or on the surface of a material.
Answer:
The size of the force developing inside the steel rod is 32039.28 N
Explanation:
Given;
length of the steel rod, L = 1.55 m
cross sectional area of the steel, A = 4.89 cm²
temperature change, ∆T = 28.0 K
coefficient of linear expansion for steel, α = 1.17 × 10⁻⁵ 1/K
Young modulus of steel, E = 200.0 GPa.
Extension of the steel is given as;
α ∆T L = FL / AE
α ∆T = F/AE
F = AEα ∆T
F = ( 4.89 x 10⁻⁴)(200 x 10⁹)(1.17 × 10⁻⁵)(28.0 K)
F = 32039.28 N
Therefore, the size of the force developing inside the steel rod when its temperature is raised, is 32039.28 N
