Answer:
1] 8500000 = <u>8.5 × 10⁶</u>
2] .000072 = <u>7.2 × 10⁻⁵</u>
3] 5.3 × 10⁴ = <u>53000</u>
4] 2.8 × 10⁻³ = <u>0.0028</u>
5] Velocity = 
V = 
<u>V = 5 m/s</u>
6] Acceleration = 
A = 
A = 
<u>A = 3 m/s²</u>
The speed with which Jennifer climb up the tree is 0.45 m/s
The distance is 10m
Time is 22 seconds
Required speed is 10/22=0.45 m/s
- Distance is the total length travelled by an object.
- Speed is distance upon time.
- It is a scalar quantity and doesn't has magnitude and direction.
- Distance traveled is the total length of the path traveled between two positions
- Distance is the total movement of an object without any regard to direction.
- Distance is the total movement of an object without any regard to direction.
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Answer:
The orbital speed of the electron is 2.296 x 10⁶ m/s
Explanation:
Given;
radius of the circular orbit, r = 5.041 × 10⁻¹¹ m
In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;

where;
h is Planck's constant
m is mass of electron
r is the radius of the circular orbit
n is the energy level of hydrogen in ground state
Substitute in these values and solve for V

Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s
To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is
KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f
, since initially cylinder was at rest
since at the ground potential energy is zero
The mathematical values are,

Here,
m = mass
g= Gravity
h = Height
V = Velocity
moment of Inertia in terms of its mass and radius
Angular velocity in terms of tangential velocity and its radius
Replacing the values we have that
mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2
gh = \frac{v^2}{2}+\frac{v^2}{4}
v = \sqrt{\frac{4gh}{3}}
From trigonometry the vertical height of inclined plane is the length of this plane for
, then


Replacing,


Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s