A) gneiss
Explanation:
The likely rock that would be found here is gneiss.
A rock type that forms under a great deal of temperature and pressure is a metamorphic rock.
Such rocks would have their minerals crystallized when subjected to these extreme conditions.
- Metamorphic rocks are changed rocks that are formed when minerals in rocks are subjected to extreme pressure and temperature before they start melting.
- These rock type can form from either sedimentary or igneous rocks. In fact a metamorphic rock can also change into another one.
- Examples of metamorphic rocks are gneiss, schist, hornfels, marble etc.
Halite, limestone and sandstone are examples of sedimentary rocks that forms by the accumulation of sediments or precipitation of ions form solutions.
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Answer:
F = -307.4 N
Explanation:
It is given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 60 m/s
Final speed of the baseball, 
Time of contact, 
(a) It is assumed to find the horizontal component of average force. It is given by :
F = -307.4 N
So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.
Answer:
B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.
Explanation:
In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential
so according to the given situation, the option b is correct
And, the rest of the options are wrong
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
