Answer:
2.74
Explanation:
Magnification = image distance/object distance
Mag = v/u
Given
v = 46cm
u = 16.8
Magnification = 46/16.8
Magnification = 2.74
Hence the magnification is 2.74
Answer:
= 3521m/s
The tangential speed is approximately 3500 m/s.
Explanation:
F = m * v² ÷ r
Fg = (G * M * m) ÷ r²
(m v²) / r = (G * M * m) / r²
v² = (G * M) / r
v = √( G * M ÷ r)
G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴
r = 32000km = 32 * 10⁶ meters
G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶
v = √1.24 * 10⁷
v = 3521.36m/s
The tangential speed is approximately 3500 m/s.
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
The answer for the following problem is mentioned below.
The option for the question is "A" approximately.
- <u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
Explanation:
Given:
Spring constant (k) = 240 N/m
amount of the compression (x) = 0.40 m
To calculate:
Elastic potential energy (E)
We know;
<em>According to the formula;</em>
E = 
× k × x × x
<u>E = </u><u> × k ×(x)²</u>
where;
E represents the elastic potential energy
K represents the spring constant
x represents amount of the compression in the string
So therefore,
Substituting the values in the above formula;
E = × 240 × (0.40)²
E = × 240 × 0.16
E = × 38.4
E = 19.2 J or approximately 20 J
<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
I have honestly never read anything about a car being disseminated,
or any instructions on how to do it, or any of the theory behind it.
