Answer:
Compression is the answer
Explanation:
got it right
Answer: 0.091 m
Explanation:
r = 1/B * √(2mV/e), where
r = radius of their circular path
B = magnitude of magnetic field = 1.29 T
m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg
V = potential difference = 2.9 kV
e = charge of the Uranium -238 ion = 1.6*10^-19 C
r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]
r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)
r = 1/1.29 * √0.0138
r = 1/1.29 * 0.117
r = 0.091 m
Therefore, the radius of their circular path is 0.091 m
Explanation:
In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:
H = V²Sin²θ/2g
where,
V = Launch Speed = 18 m/s
θ = Launch Angle = 40°
g = 9.8 m/s²
Therefore,
H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)
H = 6.83 m
Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).
Therefore, the keys will make it back to John.
We know, Potential Energy = m * g * h
Here, mass & gravity would be same, but their height will change so it will be:
ΔU = U₂ - U₁
ΔU = mgh₂ - mgh₁
ΔU = mg (h₂ - h₁)
Hope this helps!
To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:
![KE = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Where,
m = Mass
v = Velocity
Replacing we have that the Total Kinetic Energy is
![KE = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
![KE = \frac{1}{2} (5*10^{-3})(300)^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%285%2A10%5E%7B-3%7D%29%28300%29%5E2)
![KE = 225J](https://tex.z-dn.net/?f=KE%20%3D%20%20225J)
On the other hand the required Energy to heat up t melting point is
![Q_1 = mC_p \Delta T](https://tex.z-dn.net/?f=Q_1%20%3D%20mC_p%20%5CDelta%20T)
![Q_2 = L_f m](https://tex.z-dn.net/?f=Q_2%20%3D%20L_f%20m)
Where,
m = Mass
Specific Heat
Change at temperature
Latent heat of fussion
Heat required to heat up to melting point,
![Q = Q_1+Q_2](https://tex.z-dn.net/?f=Q%20%3D%20Q_1%2BQ_2)
![Q = mC_p \Delta T+L_f m](https://tex.z-dn.net/?f=Q%20%3D%20mC_p%20%5CDelta%20T%2BL_f%20m)
![Q = 5*0.128*(327-20) + 5*24.7](https://tex.z-dn.net/?f=Q%20%3D%205%2A0.128%2A%28327-20%29%20%2B%205%2A24.7)
![Q = 310J](https://tex.z-dn.net/?f=Q%20%3D%20310J)
The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.