Both balls hit the ground at the same time.
We've known this fact for about 500 years (since Galileo),
and we've known WHY for about 300 years (since Newton).
Answer:
19.0 m/s
Explanation:
First of all, we can find the acceleration of the object by using Newton's second law:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
where
F = 200 N is the net force applied to the object
m = 50.0 kg is the mass of the object
a is the acceleration
Solving for a,
![a=\frac{F}{m}=\frac{200 N}{50.0 kg}=4.0 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7B200%20N%7D%7B50.0%20kg%7D%3D4.0%20m%2Fs%5E2)
Now we can find the final speed of the object:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where
u = 3.0 m/s is the initial speed
a = 4.0 m/s^2 is the acceleration
t = 4.0 s is the time
Substituting,
![v=3.0 m/s + (4.0 m/s^2)(4.0 s)=19.0 m/s](https://tex.z-dn.net/?f=v%3D3.0%20m%2Fs%20%2B%20%284.0%20m%2Fs%5E2%29%284.0%20s%29%3D19.0%20m%2Fs)
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:
Alcohol thermometers are used rather than mercury thermometers in very cold regions because alcohol has a lower freezing point than mercury. Pure ethanol freezes at -115 degrees C, while mercury freezes at -38 degrees C.
Explanation:
It is less durable because alcohol evaporates faster than mercury. It cannot measure high temperature because of a low boiling point. It wets the walls of the thermometer, which can adversely impact the accuracy of readings
Answer:
a) Electric potential = 853 V
b) Electron speed at point B, if at Point A, the speed were zero = 1.732 × 10⁷ m/s
Explanation:
For an electron moving in an electric field with potential V,
Work done = qV where q is the charge on the electron
And the Work done is equal to the change in kinetic energy of the electron
qV = m(v₂² - v₁²)/2
V = m(v₂² - v₁²)/2q
q = 1.602 × 10⁻¹⁹C
m = 9.11 × 10⁻³¹ kg
v₁= 10⁷ m/s
v₂ = 2 × 10⁷ m/s
Putting these values in for the variables and solving
V = 853 V
b) If the electron started from rest,
qV = mv²/2
v = √(2qV/m) =√((2 × (1.602 × 10⁻¹⁹) × 853)/(9.11 × 10⁻³¹)) = 1.732 × 10⁷ m/s