Alistar places a 6.4 kg block on an incline plane that is 23 degrees above the horizontal. The block starts from rest, and in 2.
1 answer:
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Answer:
7.468 kN
Explanation:
Here the force is given in Newton
Some of the prefixes of the SI units are
kilo = 10³
Mega = 10⁶
Giga = 10⁹
Tera = 10¹²
The number is 7468.0
Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.
1 kilonewton = 1000 Newton
So, 7468 N = 7.468 kN
Answer:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:
Solving for the force F, we obtain that:
Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:
From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:
In this case, the force from the rocket's thrusters is equal to 118,000N.
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.