1 astronomical unit (1AU) or 150 000 000 km's (1.5 x<em>10^9</em> km). In SI units this is 10^12 meters
(a) 154.5 N
Let's divide the motion of the sprinter in two parts:
- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration
for a total time
During this part of the motion, he covers a distance equal to
, until he finally reaches a velocity of
. We can use the following suvat equation:
![s_1 = u t_1 + \frac{1}{2}a_1t_1^2](https://tex.z-dn.net/?f=s_1%20%3D%20u%20t_1%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_1t_1%5E2)
which reduces to
(1)
since u = 0.
- In the second part, he continues with constant speed
, covering a distance of
in a time
. This part of the motion is a uniform motion, so we can use the equation
(2)
We also know that the total time is 10.0 s, so
![t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)](https://tex.z-dn.net/?f=t_1%20%2B%20t_2%20%3D%2010.0%20s%5C%5Ct_2%20%3D%20%2810.0-t_1%29)
Therefore substituting into the 2nd equation
![s_2 = a_1 t_1 (10-t_1)](https://tex.z-dn.net/?f=s_2%20%3D%20a_1%20t_1%20%2810-t_1%29)
From eq.(1) we find
(3)
And substituting into (2)
![s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1](https://tex.z-dn.net/?f=s_2%20%3D%20%5Cfrac%7B2s_1%7D%7Bt_1%5E2%7Dt_1%20%2810-t_1%29%3D%5Cfrac%7B2s_1%7D%7Bt_1%7D%2810-t_1%29%3D%5Cfrac%7B20%20s_1%7D%7Bt_1%7D-2s_1)
Solving for t,
![s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s](https://tex.z-dn.net/?f=s_2%2B2s_1%3D%5Cfrac%7B20%20s_1%7D%7Bt_1%7D%5C%5Ct_1%20%3D%20%5Cfrac%7B20s_1%7D%7Bs_2%2B2s_1%7D%3D%5Cfrac%7B20%2845%29%7D%7B55%2B2%2845%29%7D%3D6.2%20s)
So from (3) we find the acceleration in the first phase:
And so the average force exerted on the sprinter is
![F=ma=(66 kg)(2.34 m/s^2)=154.5 N](https://tex.z-dn.net/?f=F%3Dma%3D%2866%20kg%29%282.34%20m%2Fs%5E2%29%3D154.5%20N)
b) 14.5 m/s
The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation
![v_1 = u +a_1 t_1](https://tex.z-dn.net/?f=v_1%20%3D%20u%20%2Ba_1%20t_1)
where we have
u = 0
is the acceleration
is the time of the first part
Solving the equation,
![v_1 = 0 +(2.34)(6.2)=14.5 m/s](https://tex.z-dn.net/?f=v_1%20%3D%200%20%2B%282.34%29%286.2%29%3D14.5%20m%2Fs)
Answer:
C. contains negatively charged particles
Explanation:
- J.J Thomson placed two opposite charged electric plates around a cathode ray tube which is a sealed glass tube with high vacuum
- He found that a beam of negatively charged particles are generated in the tube which got attracted to the positively charged plate.
- These subatomic particles are called electrons and they have negative charge.
Answer:
recoil velocity of canon is 0.1039 m/s
Explanation:
given,
mass of shell = 97 kg
speed at which the shell is fired = 105 m/s
angle at which the shell is fired (θ) = 60°
mass of cannon plus car = 4.9 × 10⁴ kg
by conservation of momentum
∑ P = 0
![P_{initial} +P_{final} = 0](https://tex.z-dn.net/?f=P_%7Binitial%7D%20%2BP_%7Bfinal%7D%20%3D%200)
![m_{cannon}v_{cannon} +m_{shell}v_{shell} = 0](https://tex.z-dn.net/?f=m_%7Bcannon%7Dv_%7Bcannon%7D%20%2Bm_%7Bshell%7Dv_%7Bshell%7D%20%3D%200)
![v_{cannon}=\dfrac{m_{shell}v_{shell} cos\theta}{m_{cannon}}](https://tex.z-dn.net/?f=v_%7Bcannon%7D%3D%5Cdfrac%7Bm_%7Bshell%7Dv_%7Bshell%7D%20cos%5Ctheta%7D%7Bm_%7Bcannon%7D%7D)
![v_{cannon}=\dfrac{97\times 105\times cos 60^0}{4.9\times 10^4}](https://tex.z-dn.net/?f=v_%7Bcannon%7D%3D%5Cdfrac%7B97%5Ctimes%20105%5Ctimes%20cos%2060%5E0%7D%7B4.9%5Ctimes%2010%5E4%7D)
![v_{cannon} = 0.1039\ m/s](https://tex.z-dn.net/?f=v_%7Bcannon%7D%20%3D%200.1039%5C%20m%2Fs)
recoil velocity of canon is 0.1039 m/s
Answer:
a). 1.28333 seconds
b). 186.66 seconds
Explanation:
a). Given :
Distance between the earth and the moon, d =
m
Speed of the radio waves, c =
m/s
Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :
![$t=\frac{d}{c}$](https://tex.z-dn.net/?f=%24t%3D%5Cfrac%7Bd%7D%7Bc%7D%24)
![$=\frac{3.85 \times 10^8}{3 \times 10^8}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B3.85%20%5Ctimes%2010%5E8%7D%7B3%20%5Ctimes%2010%5E8%7D%24)
= 1.28333 seconds
b). Distance between Mars and the earth, d =
m
Speed of the radio waves, c =
m/s
So, the time required for his voice to reach earth is :
![$t=\frac{d}{c}$](https://tex.z-dn.net/?f=%24t%3D%5Cfrac%7Bd%7D%7Bc%7D%24)
![$=\frac{5.6 \times 10^{10}}{3 \times 10^8}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B5.6%20%5Ctimes%2010%5E%7B10%7D%7D%7B3%20%5Ctimes%2010%5E8%7D%24)
= 186.66 seconds