Answer:
V0=27.4 m/s; t=0.8 s
Explanation:
Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:
We solve for initial speed
Now, using the same expression we estimated time to first reach 18.5 m :
Second order equation with solutions
t1=0.8 s and t2=4.8 s
The first time corresponds to the first reach.
Answer :
The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.
Explanation:
Given,
Atomic mass of silver = 107.87 g/mol
Density of silver = 10.35 g/cm^3
Converting to g/m^3,
= 10.35 g/cm^3 × 10^6cm^3/m^3
= 10.35 × 10^6 g/m^3
Avogadro's number = 6.022 × 10^23 atoms/mol
Fraction of lattice sites that are vacant in silver = 1 × 10^-6
Nag = (Na * Da)/Aag
Where,
Nag = Total number of lattice sites in Ag
Na = Avogadro's number
Da = Density of silver
Aag = Atomic weight of silver
= (6.022 × 10^23 × (10.35 × 10^6)/107.87
= 5.778 × 10^28 atoms/m^3
The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6
= 5.778 × 10^22/m^3.
Answer:
The mass of Laura and the sled combined is 887.5 kg
Explanation:
The total force due to weight of Laura and friction on the sled can be calculated as follows;
= (400 + 310) N
= 710 N
From Newton's second law of motion, "the rate of change of momentum is directly proportional to the applied force.
where;
is mass of Laura and
is mass of sled
Mass of Laura and the sled combined is calculated as follows;
given
V = Δv = 4-0 = 4m/s
t = 5 s
Therefore, the mass of Laura and the sled combined is 887.5 kg
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
