Answer:
Icy roads
Explanation:
There is so little friction you slide on it way more than other roads. :)
Answer:
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
(3)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
First, we calculate for the weight of the object by multiplying the given mass by the acceleration due to gravity which is equal to 9.8 m/s²
Weight = (14 kg)(9.8 m/s²)
Weight = 137.2 N
The component of the weight that is along the surface of the inclined plane is equal to this weight times the sine of the given angle.
Weight = (137.2 N)(sin 52°)
weight = 108.1 N
Explanation:
Given that,
Terminal voltage = 3.200 V
Internal resistance ![r= 5.00\ \Omega](https://tex.z-dn.net/?f=r%3D%205.00%5C%20%5COmega)
(a). We need to calculate the current
Using rule of loop
![E-IR-Ir=0](https://tex.z-dn.net/?f=E-IR-Ir%3D0)
![I=\dfrac{E}{R+r}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BE%7D%7BR%2Br%7D)
Where, E = emf
R = resistance
r = internal resistance
Put the value into the formula
![I=\dfrac{3.200}{1.00\times10^{3}+5.00}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B3.200%7D%7B1.00%5Ctimes10%5E%7B3%7D%2B5.00%7D)
![I=3.184\times10^{-3}\ A](https://tex.z-dn.net/?f=I%3D3.184%5Ctimes10%5E%7B-3%7D%5C%20A)
(b). We need to calculate the terminal voltage
Using formula of terminal voltage
![V=E-Ir](https://tex.z-dn.net/?f=V%3DE-Ir)
Where, V = terminal voltage
I = current
r = internal resistance
Put the value into the formula
![V=3.200-3.184\times10^{-3}\times5.00](https://tex.z-dn.net/?f=V%3D3.200-3.184%5Ctimes10%5E%7B-3%7D%5Ctimes5.00)
![V=3.18\ V](https://tex.z-dn.net/?f=V%3D3.18%5C%20V)
(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf
![\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }](https://tex.z-dn.net/?f=%5Cdfrac%7BTerminal%5C%20voltage%7D%7Bemf%7D%3D%5Cdfrac%7B3.18%7D%7B3.200%20%7D)
![\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}](https://tex.z-dn.net/?f=%5Cdfrac%7BTerminal%5C%20voltage%7D%7Bemf%7D%3D%20%5Cdfrac%7B159%7D%7B160%7D)
Hence, This is the required solution.
For conservation of energy we have to:
mgH=mv²/2
Clearing
<span> v=sqrt(2gH)
Then, by definition
</span><span> F=Δp/Δt= Δ(mv)/ Δt=m Δ(v)/Δt=
</span> =m[sqrt(2gH)-0]/Δt= m[sqrt(2gH)]/ Δt
the answer is
F=m[sqrt(2gH)]/ Δt