Answer:
Initial velocity (u) = 40 m/s
Distance travel in last 5 seconds = 100 m
Explanation:
Given:
Acceleration (a) = 8 m/s²
Final velocity (v) = 0 m/s
Find;
1] Initial velocity before 5 sec
2] Distance travel in last 5 seconds
Computation:
1] Initial velocity before 5 sec
v = u + at
0 = u + (-8)(5)
u - 40 = 0
Initial velocity (u) = 40 m/s
2] Distance travel in last 5 seconds
s = ut + (1/2)(a)(t²)
s = (40)(5) + (1/2)(-8)(5²)
s = 200 - 100
Distance travel in last 5 seconds = 100 m
<span>The
heavier the body is, the stronger its gravitational pull. Example, the Milky Way
Galaxy has a gravitational pull because of the heavenly bodies such as stars and planets are surrounding it. A strong force is exerted if the mass of another body is bigger than the other body.</span>
Using
KE = ½mv² = ½×1500×19×19 = 270750 joules
Answer:
science
Are rocks considered valuable natural resources?(1 point)
No, rocks are not valuable because they contain a lot of waste.
No, rocks are not valuable because they are nonliving solids.
Yes, rocks are valuable because they are useful to people.
Yes, rocks are valuable because they are part of Earth’s crust.
Aluminum is produced from bauxite ore and can be used for many things. Which option describes a common production use of aluminum?(1 point)
steel objects
electrical wiring and pipes
fuel for nuclear power plants
lightweight containers and drugs
What are common uses of the mineral quartz?(1 point)
building and paving materials
glass manufacturing and paints
automobile manufacturing and packaging
iron and steel production
Bornite is which type of ore?(1 point)
copper ore
iron ore
bauxite ore
uraninite ore
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ = σ ( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23