Answer:
0.2289
Explanation:
Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then
P= 700*9.81*2.5=17167.5 W= 17.1675 kW
To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%
<h2>
Answer: x=125m, y=48.308m</h2>
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:
x-component:
(1)
Where:
is the projectile's initial speed
is the angle
is the time since the projectile is launched until it strikes the target
is the final horizontal position of the projectile (the value we want to find)
y-component:
(2)
Where:
is the initial height of the projectile (we are told it was launched at ground level)
is the final height of the projectile (the value we want to find)
is the acceleration due gravity
Having this clear, let's begin with x (1):
(3)
(4) This is the horizontal final position of the projectile
For y (2):
(5)
(6) This is the vertical final position of the projectile
The answer is asthenosphere
Answer:
Compound.
Explanation:
A compound is a substance formed when two or more elements are chemically joined. Water, salt, and sugar are examples of compounds. When the elements are joined, the atoms lose their individual properties and have different properties from the elements they are composed of.
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux
![\phi=NBA](https://tex.z-dn.net/?f=%5Cphi%3DNBA)
Put the value into the formula
![\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2](https://tex.z-dn.net/?f=%5Cphi%3D10%5Ctimes1.0%5Ctimes10%5E%7B-3%7D%5Ctimes%5Cpi%5Ctimes%280.5%5Ctimes10%5E%7B-2%7D%29%5E2)
![\phi=7.85\times10^{-7}\ Tm^2](https://tex.z-dn.net/?f=%5Cphi%3D7.85%5Ctimes10%5E%7B-7%7D%5C%20Tm%5E2)
We need to calculate the induced emf
Using formula of induced emf
![\epsilon=\dfrac{d\phi}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3D%5Cdfrac%7Bd%5Cphi%7D%7Bdt%7D)
Put the value into the formula
![\epsilon=\dfrac{7.85\times10^{-7}}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7Bdt%7D)
Put the value of emf from ohm's law
![\epsilon =IR](https://tex.z-dn.net/?f=%5Cepsilon%20%3DIR)
![IR=\dfrac{7.85\times10^{-7}}{dt}](https://tex.z-dn.net/?f=IR%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7Bdt%7D)
![Idt=\dfrac{7.85\times10^{-7}}{R}](https://tex.z-dn.net/?f=Idt%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7BR%7D)
![Idt=\dfrac{7.85\times10^{-7}}{0.50}](https://tex.z-dn.net/?f=Idt%3D%5Cdfrac%7B7.85%5Ctimes10%5E%7B-7%7D%7D%7B0.50%7D)
![Idt=0.00000157=1.57\times10^{-6}\ C](https://tex.z-dn.net/?f=Idt%3D0.00000157%3D1.57%5Ctimes10%5E%7B-6%7D%5C%20C)
We know that,
![Idt=dq](https://tex.z-dn.net/?f=Idt%3Ddq)
![dq=1.57\times10^{-6}\ C](https://tex.z-dn.net/?f=dq%3D1.57%5Ctimes10%5E%7B-6%7D%5C%20C)
We need to calculate the voltage across the capacitor
Using formula of charge
![dq=C dV](https://tex.z-dn.net/?f=dq%3DC%20dV)
![dV=\dfrac{dq}{C}](https://tex.z-dn.net/?f=dV%3D%5Cdfrac%7Bdq%7D%7BC%7D)
Put the value into the formula
![dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}](https://tex.z-dn.net/?f=dV%3D%5Cdfrac%7B1.57%5Ctimes10%5E%7B-6%7D%7D%7B1.0%5Ctimes10%5E%7B-6%7D%7D)
![dV=1.57\ V](https://tex.z-dn.net/?f=dV%3D1.57%5C%20V)
Hence, The voltage across the capacitor is 1.57 V.