Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components
![(-3\cdot (-6), -2 \cdot (-6))=(18,12)](https://tex.z-dn.net/?f=%28-3%5Ccdot%20%28-6%29%2C%20-2%20%5Ccdot%20%28-6%29%29%3D%2818%2C12%29)
The magnitude is given by Pythagorean's theorem:
![m=\sqrt{18^2+12^2}=21.6](https://tex.z-dn.net/?f=m%3D%5Csqrt%7B18%5E2%2B12%5E2%7D%3D21.6)
and the direction is given by the arctan of the ratio between the y-component and the x-component:
![\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B12%7D%7B18%7D%29%3D33.7%5E%7B%5Ccirc%7D)
Answer:
After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+
)
Explanation:
I believe it is a part C question.
The derivative of V and P will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........
Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.
Answer:
The answer to your question is Alpha particles.
Explanation: An electron released by a radioactive nucleus that causes a neutron to change into a proton is called a beta particle.
Answer:
![\frac{T_t}{T_c} = 1.32](https://tex.z-dn.net/?f=%5Cfrac%7BT_t%7D%7BT_c%7D%20%3D%201.32)
Explanation:
The torque applied on an object can be calculated by the following formula:
![T = Fr](https://tex.z-dn.net/?f=T%20%3D%20Fr)
where,
T = Torque
F = Applied Force
r = radius of the wheel
For car wheel:
![T_c = Fr_c\\](https://tex.z-dn.net/?f=T_c%20%3D%20Fr_c%5C%5C)
For truck wheel:
![T_t = Fr_t](https://tex.z-dn.net/?f=T_t%20%3D%20Fr_t)
Dividing both:
![\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}](https://tex.z-dn.net/?f=%5Cfrac%7BT_t%7D%7BT_c%7D%20%3D%20%5Cfrac%7BFr_t%7D%7BFr_c%7D)
for the same force applied on both wheels:
![\frac{T_t}{T_c} = \frac{r_t}{r_c} \\](https://tex.z-dn.net/?f=%5Cfrac%7BT_t%7D%7BT_c%7D%20%3D%20%5Cfrac%7Br_t%7D%7Br_c%7D%20%5C%5C)
where,
rt = radius of the truck steering wheel = 0.25 m
rc = radius of the car steering wheel = 0.19 m
Therefore,
![\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\](https://tex.z-dn.net/?f=%5Cfrac%7BT_t%7D%7BT_c%7D%20%3D%20%5Cfrac%7B0.25%5C%20m%7D%7B0.19%5C%20m%7D%20%5C%5C)
![\frac{T_t}{T_c} = 1.32](https://tex.z-dn.net/?f=%5Cfrac%7BT_t%7D%7BT_c%7D%20%3D%201.32)
Answer:
5.38 m/s
Explanation:
Given (in the x direction):
Δx = 2.45 m
v₀ = v cos 42.5°
a = 0 m/s²
Δx = v₀ t + ½ at²
(2.45 m) = (v cos 42.5°) t + ½ (0 m/s²) t²
2.45 = (v cos 42.5°) t
t = 3.32 / v
Given (in the y direction):
Δy = 0.373 m
v₀ = v sin 42.5°
a = -9.8 m/s²
Δx = v₀ t + ½ at²
(0.373 m) = (v sin 42.5°) t + ½ (-9.81 m/s²) t²
0.373 = (v sin 42.5°) t − 4.905 t²
0.373 = (v sin 42.5°) (3.32 / v) − 4.905 (3.32 / v)²
0.373 = 2.25 − 54.2 / v²
v = 5.38
Graph:
desmos.com/calculator/5n30oxqmuu