So we want to know what is the distance d of the object from the lens if the height of the object is h=6 cm, focal length of the lens is f=5 cm and the distance d=15 cm is the distance of the object from the lens. From the formula for the convex lens 1/f=(1/D + 1/d) where D is the distance of the image from the lens we can get D after solving for D: 1/D=(1/f) - (1/d),
1/D=(1/5)-(1/15)=0.2-0,06667=0.13333 so f=1/0.13333=7.500187 cm. Rounded to the nearest hundredth D=7.50 cm. That is very close to 7.69 cm so the correct answer is the third one.
Fossil fuels like coal for instance , when they are provided with heat energy (heating them) they start producing Kinetic energy . This kinetic energy will be passed along a turbine which transforms this K.E to an electrical energy used in different useful forms.
Answer:
Mass = 873.6kg = 0.874 × 10^3 kg
Order = 3
Question:
Compute the order of magnitude of the mass of a bathtub half full of pennies.(Assume the pennies are made entirely of copper and the tub measures 1.3 m by 0.5 m by 0.3 m)
Explanation:
Given:
Dimensions of the tub = 1.3 m by 0.5 m by 0.3 m
Density of copper d = 8960kg/m^3
Mass = volume × density
Volume of tub = L×B×H = 1.3 × 0.5 × 0.3 = 0.195m^3
Since the tub is half full:
V = 0.195/2 = 0.0975m^3
Mass = 8960kg/m^3 × 0.0975m^3
Mass = 873.6kg = 0.874 × 10^3 kg
Order = 3
Note: Order of magnitude is of the form:
N = a × 10^b
Where;
b = order of magnitude
and
1/√10 </= a < √10
Answer:
a) t =12[s]; b) x = 348[m]
Explanation:
We can solve this problem using the following kinematics equations:
a)
where:
vf = final velocity = 12 [m/s]
vo= initial velocity = 6 [m/s]
a = acceleration = 0.5[m/s^2]
t = time [s]
Now clearing the time t, we have:
![t=\frac{v_{f} -v_{o} }{a} \\t = \frac{12-6}{0.5} \\t=12[s]](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv_%7Bf%7D%20-v_%7Bo%7D%20%7D%7Ba%7D%20%5C%5Ct%20%3D%20%5Cfrac%7B12-6%7D%7B0.5%7D%20%5C%5Ct%3D12%5Bs%5D)
b)
We can calculate the displacement for the first 12 [s] then using the equation for the constant velocity we can calculate the other displacement for the 20[s].
![v_{f}^{2}= v_{o}^{2}+2*a*x_{1} \\therefore\\x_{1} =\frac{v_{f}^{2}-v_{o}^{2}}{2*a} \\x_{1} =\frac{12^{2}-6^{2}}{2*.5}\\x_{1} =108[m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20%20v_%7Bo%7D%5E%7B2%7D%2B2%2Aa%2Ax_%7B1%7D%20%5C%5Ctherefore%5C%5Cx_%7B1%7D%20%3D%5Cfrac%7Bv_%7Bf%7D%5E%7B2%7D-v_%7Bo%7D%5E%7B2%7D%7D%7B2%2Aa%7D%20%5C%5Cx_%7B1%7D%20%3D%5Cfrac%7B12%5E%7B2%7D-6%5E%7B2%7D%7D%7B2%2A.5%7D%5C%5Cx_%7B1%7D%20%3D108%5Bm%5D)
The we can calculate the second displacement for the constant velocity:
![x_{2} =x_{o}+v*t_{2} \\ x_{2} =0+12*(20)\\x_{2} =240[m]](https://tex.z-dn.net/?f=x_%7B2%7D%20%3Dx_%7Bo%7D%2Bv%2At_%7B2%7D%20%5C%5C%20x_%7B2%7D%20%20%3D0%2B12%2A%2820%29%5C%5Cx_%7B2%7D%20%3D240%5Bm%5D)
x = x1 + x2
x = 108 + 240
x = 348[m]
Then only the component of the force that's parallel to the displacement
is used to calculate the work. The component of force that's perpendicular
to the displacement doesn't move through any distance at all, so its contribution
to the total work is zero.
