Answer:
(a). The weight of spacecraft at height 6450 km is
.
(b). The weight of spacecraft at height 33700 km is 182.46 N.
Explanation:
Given that,
Radius of earth = 6450 km
Weight of spacecraft = 7070 N
We need to calculate the weight of the spacecraft at a height 6450 km above Earth’s surface
Using formula of weight
![W_{r}=W(\dfrac{r_{e}}{r_{e}+h})^2](https://tex.z-dn.net/?f=W_%7Br%7D%3DW%28%5Cdfrac%7Br_%7Be%7D%7D%7Br_%7Be%7D%2Bh%7D%29%5E2)
Where, W = weight of spacecraft
h = height
Put the value into the formula
![W_{r}=7070\times(\dfrac{6450\times10^{3}}{6450\times10^3+6450\times10^3})^2](https://tex.z-dn.net/?f=W_%7Br%7D%3D7070%5Ctimes%28%5Cdfrac%7B6450%5Ctimes10%5E%7B3%7D%7D%7B6450%5Ctimes10%5E3%2B6450%5Ctimes10%5E3%7D%29%5E2)
![W_{r}=1.767\times10^{3}\ N](https://tex.z-dn.net/?f=W_%7Br%7D%3D1.767%5Ctimes10%5E%7B3%7D%5C%20N)
(b). We need to calculate the weight of spacecraft at height 33700 km
Using formula of weight
![W_{r}=7070\times(\dfrac{6450\times10^{3}}{6450\times10^3+33700\times10^3})^2](https://tex.z-dn.net/?f=W_%7Br%7D%3D7070%5Ctimes%28%5Cdfrac%7B6450%5Ctimes10%5E%7B3%7D%7D%7B6450%5Ctimes10%5E3%2B33700%5Ctimes10%5E3%7D%29%5E2)
![W_{r}=182.46\ N](https://tex.z-dn.net/?f=W_%7Br%7D%3D182.46%5C%20N)
Hence, (a). The weight of spacecraft at height 6450 km is
.
(b). The weight of spacecraft at height 33700 km is 182.46 N.
The series circuit is the one in which the current flowing through each bulb will be the same as the current at point x.
<h3>What is a series circuit?</h3>
When bulbs are connected in series, the bulbs are connected in an end to end manner. The same current flows through all the bulbs when they are connected in series.
As such, the series circuit is the one in which the current flowing through each bulb will be the same as the current at point x.
Learn more about series circuit:brainly.com/question/11409042?
#SPJ1
Heya!!
For calculate final velocity, lets applicate formula
![\boxed{V=V_o+a*t}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3DV_o%2Ba%2At%7D)
<u>Δ Being Δ</u>
V = Final Velocity = ?
Vo = Initial velocity = 0 m/s
a = Aceleration = 5 m/s²
t = Time = 12 s
⇒ Let's replace according the formula:
![\boxed{V=0\ m/s +5\ m/s*12\ s}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3D0%5C%20m%2Fs%20%2B5%5C%20m%2Fs%2A12%5C%20s%7D)
⇒ Resolving
![\boxed{V=60\ m/s}](https://tex.z-dn.net/?f=%5Cboxed%7BV%3D60%5C%20m%2Fs%7D)
Result:
The velocity after 10 sec is <u>60 meters per second (m/s)</u>
Good Luck!!
We use the voltage division problem between the load resistance, amplifier output resistance as
.
Here,
is the output voltage,
is the amplifier voltage,
is the load resistance and
is the amplifier output resistance.
Therefore,
.
Thus, the amplifier output resistance is
.
Answer:
![F_0 = 393 N](https://tex.z-dn.net/?f=F_0%20%3D%20393%20N)
Explanation:
As we know that amplitude of forced oscillation is given as
![A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BF_0%7D%7B%20m%28%5Comega%5E2%20-%20%5Comega_0%5E2%29%7D)
here we know that natural frequency of the oscillation is given as
![\omega_0 = \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega_0%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
here mass of the object is given as
![m = \frac{W}{g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7BW%7D%7Bg%7D)
![\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}](https://tex.z-dn.net/?f=%5Comega_0%20%3D%20%5Csqrt%7B%5Cfrac%7B220%7D%7B%5Cfrac%7B30%7D%7B9.81%7D%7D%7D)
![\omega_0 = 8.48 rad/s](https://tex.z-dn.net/?f=%5Comega_0%20%3D%208.48%20rad%2Fs)
angular frequency of applied force is given as
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
![\omega = 2\pi(10.5) = 65.97 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%2810.5%29%20%3D%2065.97%20rad%2Fs)
now we have
![0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}](https://tex.z-dn.net/?f=0.03%20%3D%20%5Cfrac%7BF_0%7D%7B3.06%2865.97%5E2%20-%208.48%5E2%29%7D)
![F_0 = 393 N](https://tex.z-dn.net/?f=F_0%20%3D%20393%20N)