Term-1 physic讨论区

MichelleLow

i very weak in projectile that part, help solve them, better with diagram..
1. A ball is fired into the air at an angle 30 degree to the horizontal. If its initial velocity is 14.7 m s^-1, how long will it take to return to the ground? (g= 9.81 m s^-2)

2. A student who stands 2 m away from a vertical wall throws a stone in a horizontal direction at a height of 1.6m from the ground. The stone hits the wall at a point which is 1.2 m from the ground. Determine:-
a) the time taken by the stone to reach the wall, ans=0.28
b) the initial speed of stone, ans. =7.1
c) speed of the stone when it strike the wall, ans. = 7.6m s^-1, 21.5 degree
(g= 10m s^-2)

YetSin

MichelleLow 发表于 7-7-2014 06:46 PM
i very weak in projectile that part, help solve them, better with diagram..
1. A ball ...

1) v_y=v sin \theta =14.7 sin 30 = 7.35ms^-1
    s=ut + 1/2 at^2
    t(7.35-4.905t)=0
    t= 7.35/4.905
2)
a) u_y =0
    s=ut+1/2 at^2
    0.4=5t^2
    t=\sqrt(0.08)=0.28s
b) a_x=0
    s=v_x t
    v_x = s/t = 2/0.28 = 7.1ms^-1
c) v_y = 0.28(10)=2.8
    v=\sqrt(2.8^2+7.1^2)=7.6ms^-1
    \theta = arctan (2.8/7.1)=21.5degree

MichelleLow

YetSin 发表于 8-7-2014 11:01 AM
1) v_y=v sin \theta =14.7 sin 30 = 7.35ms^-1
    s=ut + 1/2 at^2
    t(7.35-4.905t)=0

Hello~,for no. 2 c) if i use the formula of s=ut+ 1/2 at^2
                                                          1.2=Vy(t)+1/2(10)(0.28)^2
is this equation can be used for??but my final answer of velocity and the angle are a little difference

YetSin

MichelleLow 发表于 8-7-2014 06:21 PM
Hello~,for no. 2 c) if i use the formula of s=ut+ 1/2 at^2
                                       ...

why are you using that formula? what you want are the resultant velocity which only need v_x and v_y and has nothing to do with s=ut+1/2 at^2

MichelleLow

YetSin 发表于 8-7-2014 11:39 PM
why are you using that formula? what you want are the resultant velocity which only need v_x and v ...

ok..thank you~

MichelleLow

A car of mass 1000kg is moving with a constant velocity of 20ms^-1 along a horizontal road. The output power of the car is 20kW. The output power is then suddenly increased to 60kW. What is the acceleration of the car.(Ans. =2 ms^-1)

How to solve it??

schweitzer

MichelleLow 发表于 18-7-2014 07:39 PM
A car of mass 1000kg is moving with a constant velocity of 20ms^-1 along a horizontal road. The outp ...

大概跟你讲一下。

首先，车在路上跑就会有阻力，那20kW的power 就是用来克服阻力的，因为constant velocity = 20 ， KE=mv^2 也是固定的。

如果power 突然加到60k， 那么就会有额外的40kW 可以被用来加速。

用 P = F dot v , 40k = F dot 20, F = 2k. Then, F = ma, 20k = 1000a, a = 2.

PS：麻烦lz改下标题， 物理学的英文应该是 Physics。

doeramon

为什么dynamic friction< limiting static friction?
解释什么是work done?(不要给我formula)
有static friction的object,  work done 会少过没有friction的object（就是这句才让我想要问什么是work done)

GGJY

请问谁有experiment 3 dynamics的sample?可以拍上來吗？ for calculation,discussion and conclusion...不知道要怎样写。。。:Q

媁媁

求解 谢谢....

特务47

请问这题要怎么做？

cmstrawberry97

特务47 发表于 20-5-2016 01:01 AM
请问这题要怎么做？

67.1N 用 60square加30square再square root
画三角形出来就算到了

特务47

cmstrawberry97 发表于 20-5-2016 03:44 PM
67.1N 用 60square加30square再square root
画三角形出来就算到了

那90度不用理它？

特务47

再次向各位大大请教，可以解释这题怎么做吗？