Answer:
Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas
Explanation:
<u>TEMPERATURE</u>:
Decreasing the temperature will slow down the molecules. Hence, less no. of collisions will take place between walls of object and molecules. This will result in decrease of pressure.
Therefore, the pressure of a gas can be decreased by increasing its temperature.
<u>NUMBER OF GAS PARTICLES</u>:
Decreasing the number of particles will result in less no. of collisions, hence decreasing the pressure.
Therefore, the pressure of a gas can be decreased by decreasing its no. of molecules or no. of particles.
<u>AREA OF OBJECT:</u>
The pressure is given by the formula:
![P = \frac{F}{A}\\\\For constant Force (F):\\P\ \alpha\ \frac{1}{A}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BF%7D%7BA%7D%5C%5C%5C%5CFor%20constant%20Force%20%28F%29%3A%5C%5CP%5C%20%5Calpha%5C%20%5Cfrac%7B1%7D%7BA%7D)
where,
A = Area of Object
Therefore, the pressure of a gas can be decreased by increasing area of object.
So, the correct option is:
<u>Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas</u>
Answer:
![14.98\ \text{kg m/s}](https://tex.z-dn.net/?f=14.98%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
![45.26^{\circ}](https://tex.z-dn.net/?f=45.26%5E%7B%5Ccirc%7D)
Explanation:
= Initial momentum of the pin = 13 kg m/s
= Initial momentum of the ball = 18 kg m/s
= Momentum of the ball after hit
= Angle ball makes with the horizontal after hitting the pin
= Angle the pin makes with the horizotal after getting hit by the ball
Momentum in the x direction
![P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}](https://tex.z-dn.net/?f=P_i%3DP_1%5Ccos55%5E%7B%5Ccirc%7D%2BP_2%5Ccos%5Ctheta%5C%5C%5CRightarrow%20P_2%5Ccos%5Ctheta%3DP_i-P_1%5Ccos55%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20P_2%5Ccos%5Ctheta%3D18-13%5Ccos55%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20P_2%5Ccos%5Ctheta%3D10.54%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
Momentum in the y direction
![P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}](https://tex.z-dn.net/?f=P_1%5Csin55%3DP_2%5Csin%5Ctheta%5C%5C%5CRightarrow%20P_2%5Csin%5Ctheta%3D13%5Csin55%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20P_2%5Csin%5Ctheta%3D10.64%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
![(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}](https://tex.z-dn.net/?f=%28P_2%5Ccos%5Ctheta%29%5E2%2B%28P_2%5Csin%5Ctheta%29%5E2%3DP_2%5E2%5C%5C%5CRightarrow%20P_2%3D%5Csqrt%7B10.54%5E2%2B10.64%5E2%7D%5C%5C%5CRightarrow%20P_2%3D14.98%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
The pin's resultant velocity is ![14.98\ \text{kg m/s}](https://tex.z-dn.net/?f=14.98%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
![P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}](https://tex.z-dn.net/?f=P_2%5Csin%5Ctheta%3D10.64%5C%5C%5CRightarrow%20%5Ctheta%3Dsin%5E%7B-1%7D%5Cdfrac%7B10.64%7D%7B14.98%7D%5C%5C%5CRightarrow%20%5Ctheta%3D45.26%5E%7B%5Ccirc%7D)
The pin's resultant direction is
below the horizontal or to the right.
Answer : The correct option is, X : radius and Y : tangential speed
Explanation :
Centripetal force : It is defined as a force which acts on the object to keep the object moving along a circular path.
Formula for centripetal force :
![F_c=\frac{m\times V^2}{r}](https://tex.z-dn.net/?f=F_c%3D%5Cfrac%7Bm%5Ctimes%20V%5E2%7D%7Br%7D)
where,
= centripetal force
m = mass
r = radius
V = tangential speed
The quantity belong in cell X is radius and in the cell Y is tangential speed
Hence, the correct option is, X : radius and Y : tangential speed
Answer:
Explanation:
For resistance of a wire , the formula is as follows .
R = ρ L/S
where ρ is specific resistance , L is length and S is cross sectional area of wire .
for first wire resistance
R₁ = ρ 3L/3a = ρ L/a
for second wire , resistance
R₂ = ρ 3L/6a
= .5 ρ L/a
For 3 rd wire resistance
R₃ = ρ 6L/3a
= 2ρ L/a
For fourth wire , resistance
R₄ = ρ 6L/6a
= ρ L/a
So the smallest resistance is of second wire .
Its resistance is .5 ρ L/a
There are two types of flexibility exercises: static stretching, in which you stretch a muscle without moving, and dynamic stretching, which combines stretching with movements.