Answer:3.68 kg
Explanation:
Given
mass ![m_1=3 kg](https://tex.z-dn.net/?f=m_1%3D3%20kg)
distance traveled by
is 50 cm in 1 s
using ![s=ut+\frac{at^2}{2}](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7Bat%5E2%7D%7B2%7D)
![0.50=0+\frac{a\cdot 1^2}{2}](https://tex.z-dn.net/?f=0.50%3D0%2B%5Cfrac%7Ba%5Ccdot%201%5E2%7D%7B2%7D)
![a=1 m/s^2](https://tex.z-dn.net/?f=a%3D1%20m%2Fs%5E2)
Suppose T is the tension in string and
is the mass of other body
For ![m_1](https://tex.z-dn.net/?f=m_1)
![T-m_1g=m_1a](https://tex.z-dn.net/?f=T-m_1g%3Dm_1a)
![T=m_1(g+a)](https://tex.z-dn.net/?f=T%3Dm_1%28g%2Ba%29)
for other body
![m_2g-T=m_2a](https://tex.z-dn.net/?f=m_2g-T%3Dm_2a)
![T=m_2(g-a)](https://tex.z-dn.net/?f=T%3Dm_2%28g-a%29)
Equating value of Tension
![m_2=m_1\times \frac{g+a}{g-a}](https://tex.z-dn.net/?f=m_2%3Dm_1%5Ctimes%20%5Cfrac%7Bg%2Ba%7D%7Bg-a%7D)
![m_2=3\times \frac{10.8}{8.8}](https://tex.z-dn.net/?f=m_2%3D3%5Ctimes%20%5Cfrac%7B10.8%7D%7B8.8%7D)
![m_2=3.68 kg](https://tex.z-dn.net/?f=m_2%3D3.68%20kg)
Answer:
a) ![F_{net}=0](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D0)
b) ![T=0](https://tex.z-dn.net/?f=T%3D0)
Explanation:
From the question we are told that:
Dimensions:
![L*B=22.0*35.0cm](https://tex.z-dn.net/?f=L%2AB%3D22.0%2A35.0cm)
Current ![I=1.40A](https://tex.z-dn.net/?f=I%3D1.40A)
Magnetic field ![B=1.40](https://tex.z-dn.net/?f=B%3D1.40)
Therefore
![Area=L*B](https://tex.z-dn.net/?f=Area%3DL%2AB)
![A=22.0*35.0cm](https://tex.z-dn.net/?f=A%3D22.0%2A35.0cm)
![A=770cm=>770*0^{-4}](https://tex.z-dn.net/?f=A%3D770cm%3D%3E770%2A0%5E%7B-4%7D)
a)
Generally Force on Looping gives
![F_1-F_2](https://tex.z-dn.net/?f=F_1-F_2)
![F_3=F_4](https://tex.z-dn.net/?f=F_3%3DF_4)
Therefore
![F_{net}=0](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D0)
b)
Generally the equation for Torque is mathematically given by
![T=i*Asin \theta](https://tex.z-dn.net/?f=T%3Di%2AAsin%20%5Ctheta)
Since A and B are on opposite direction
![\theta=180](https://tex.z-dn.net/?f=%5Ctheta%3D180)
Therefore
![T=1.40*770*10^{-4}sin 180](https://tex.z-dn.net/?f=T%3D1.40%2A770%2A10%5E%7B-4%7Dsin%20180)
![T=0](https://tex.z-dn.net/?f=T%3D0)
Answer
given,
mass of block = 0.21 Kg
speed = 1.70 m/s
spring constant = k = 4.50 N/m
using conservation of energy
a) K.E = P.E
![\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dkx%5E2)
![\dfrac{1}{2}\times 0.21 \times 1.7^2 = \dfrac{1}{2}\times 4.5 \times x^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%200.21%20%5Ctimes%201.7%5E2%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%204.5%20%5Ctimes%20x%5E2)
![0.1348= x^2](https://tex.z-dn.net/?f=0.1348%3D%20x%5E2)
x = 0.367 m
b) if the track is not friction less the maximum compression will be same as the compression in the part a.
Answer:
Explanation:
Let the velocity of car and truck be u and breaking acceleration be a .
We shall have to assume the reflex time of the driver of the car . By the time he applies brake , his car will cover some distance . There will be some time tag between the time the truck starts decelerating and the driver of the car responding to that . During this period the car will not start decelerating . It will keep on moving with uniform velocity of u .
Let this time lag be t .
b )
For answer see the attached file
c )
The minimum trailing distance will be the distance covered by car before it starts decelerating in response to truck's deceleration .
minimum trailing distance d = u x t
d ) u = 80 km / h = 22.22 m /s
reflex action time t = 0.1 s ( assumed time )
d = 22.22 x .1
= 2.2 m