Answer:
4. It is the force of the road on the tires (an external force) that stops the car.
Explanation:
If there is no friction between the road and the tires, the car won't stop.
You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.
The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.
Hello!
Which nuclei is NOT radioactive?
A) Am-241 B) Mg-24 C) Pu-241 D) U-238
Solving:
It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.
Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.
Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.
Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.
U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.
Answer:
B) Mg-24
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I Hope this helps, greetings ... Dexteright02! =)
Gravitational potential energy = mgh or mass times acceleration due to gravity times the height
Here the mass is 0.25kg, the height is 10m, and gravity is 9.8m/s^2 so...
GPE = (0.25)(10)(9.8)
GPE = 24.5 J
<span>In most cases, magma differentiation (a.k.a. fractional crystallization produces magma with higher silica content than the parent magma. Fractional crystallization removes early formed minerals in magma. The liquid that does not react to the process remains in the magma. </span>
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg