The first, fourth and sixth options. A person walks up a flight of stairs, wind lifts a balloon into the air, and a weightlifter holds a barbell straight overhead.
Answer:
The impulse on the object is 60Ns.
Explanation:
Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.
F = m a
F = m(
)
⇒ Ft = m(
-
)
where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object,
is its initialvelocity and
is the final velocity of the object.
Therefore,
impulse = Ft = m(
-
)
From the question, m = 3kg,
= 0m/s and
= 20m/s.
So that,
Impulse = 3 (20 - 0)
= 3(20)
= 60Ns
The impulse on the object is 60Ns.
Answer:
The height of Sears Tower is 1448.5 feet.
Explanation:
<h3>
We apply the free fall formula to the ball:
</h3><h3>
![y=v_{o} *t+\frac{1}{2} *g*t^{2}](https://tex.z-dn.net/?f=y%3Dv_%7Bo%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D)
</h3><h3>y: The vertical distance the ball moves at time t </h3><h3>
![v_{o}](https://tex.z-dn.net/?f=v_%7Bo%7D)
i: Initial speed
</h3><h3>g=Gravity acceleration=
![9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )](https://tex.z-dn.net/?f=9.8%2A%28%5Cfrac%7B%5Cfrac%7B1ft%7D%7B0.305m%7D%20%7D%7Bs%5E%7B2%7D%20%7D%20%29)
</h3>
Known information
We know that the vertical distance (y) that the ball moves in 9,5s is equal to height of Sears Tower (h).
Too we know that the ball is released from rest, then,
=0
Height of Sears Tower calculation:
We replace in the equation 1 the data following;
![y=h](https://tex.z-dn.net/?f=y%3Dh)
![v_{o} =0](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D0)
![g=32,1\frac{ft}{s^{2} }](https://tex.z-dn.net/?f=g%3D32%2C1%5Cfrac%7Bft%7D%7Bs%5E%7B2%7D%20%7D)
![t= 9,5s](https://tex.z-dn.net/?f=t%3D%209%2C5s)
![h=0*9.5+\frac{1}{2} *32.1*9.5^{2}](https://tex.z-dn.net/?f=h%3D0%2A9.5%2B%5Cfrac%7B1%7D%7B2%7D%20%2A32.1%2A9.5%5E%7B2%7D)
![h=1448.5 ft](https://tex.z-dn.net/?f=h%3D1448.5%20ft)
Answer: The height of Sears Tower is 1448.5 ft
Answer:
![\frac{M_e}{M_s} = 3.07 \times 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7BM_e%7D%7BM_s%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-6%7D)
Explanation:
As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula
![T = 2\pi \sqrt{\frac{r^3}{GM}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGM%7D%7D)
now for the time period of moon around the earth we can say
![T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}](https://tex.z-dn.net/?f=T_1%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br_1%5E3%7D%7BGM_e%7D%7D)
here we know that
![T_1 = 0.08 year](https://tex.z-dn.net/?f=T_1%20%3D%200.08%20year)
![r_1 = 0.0027 AU](https://tex.z-dn.net/?f=r_1%20%3D%200.0027%20AU)
= mass of earth
Now if the same formula is used for revolution of Earth around the sun
![T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}](https://tex.z-dn.net/?f=T_2%20%3D%202%5Cpi%5Csqrt%7B%5Cfrac%7Br_2%5E3%7D%7BGM_s%7D%7D)
here we know that
![r_2 = 1 AU](https://tex.z-dn.net/?f=r_2%20%3D%201%20AU)
![T_2 = 1 year](https://tex.z-dn.net/?f=T_2%20%3D%201%20year)
= mass of Sun
now we have
![\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_2%7D%7BT_1%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Br_2%5E3%20M_e%7D%7Br_1%5E3%20M_s%7D%7D)
![\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.08%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B1%20M_e%7D%7B%280.0027%29%5E3M_s%7D%7D)
![12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}](https://tex.z-dn.net/?f=12.5%20%3D%20%5Csqrt%7B%285.08%20%5Ctimes%2010%5E7%29%5Cfrac%7BM_e%7D%7BM_s%7D%7D)
![\frac{M_e}{M_s} = 3.07 \times 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7BM_e%7D%7BM_s%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-6%7D)