Answer:
t = 4.468 h
Explanation:
For this exercise let's start by calculating the distance where the boat is
the first trip is 50 miles at 45 °, then 90 miles at 80 °,
to find the total distance let's find the distance of each displacement
cos 45 = x₁ / 50
sin 45 = y₁ / 50
x₁ = 50 cos 45 = 35.35 miles
y₁ = 50 cos 45 = 35.35 miles
cos 80 = x₂ / 90
sin 80 = y₂ / 90
x₂ = 90 cos 80 = 15.63 miles
y₂ = 90 sin 80 = 88.63 miles
let's find the total displacement in each axis
x_total = x₁ + x₂
x_total = 35.35 + 15.63
x_total = 50.98 miles
y_total = y₁ + y₂
y_total = 35.35 + 88.63
y_total = 123.98 miles
Let's use the Pythagorean theorem to find the modulus of the displacement
R = √ (x_total² + y_total²)
R = √ (50.98² + 123.98²)
R = 134.05 miles
The boat goes at a constant speed,
v = R / t
t = R / v
let's calculate
t = 134.05 / 30
t = 4.468 h
Answer:
(a) The speed of the first particle is 1.75 m/s. The speed of the second particle is 6.9 m/s after the collision.
(b) The speed of the first particle is 3.45 m/s in the negative direction. The speed of the second particle is 1.73 m/s.
(c) The final kinetic energy of the incident particle in part (a) and part(b) is 0.0031 J and 0.011 J, respectively.
Explanation:
(a)
In an elastic collision, both momentum and energy is conserved.

Combining these equations will give the speed of the second particle.

We can use this to find the speed of the first particle.

(b)
If m_2 = 10g.


The minus sign indicates that the first particle turns back after the collision.
(c)
The final kinetic energy of the particle in part (a) and part (b) is
Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R = ![\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]](https://tex.z-dn.net/?f=%5Crho%20%28x%29%20%3D%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%20%5C%20m%7D%29%5E2%5D)
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as = 
![dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}](https://tex.z-dn.net/?f=dR%20%3D%20%5Cfrac%7B%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%7D%29%5E2%5D%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%2810%5E%7B-3%7D%29%5E2%7D)
Taking the integral of both sides;we have:
![\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5ER_0%20%20dR%20%3D%20%5Cint%5Climits%5E2_0%203.185%20%5C%20%5B1%2Bx%5E2%5D%20%5C%20dx)
![R = 3.185 [x + \frac {x^3}{3}}]^2__0](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5Bx%20%2B%20%5Cfrac%20%7Bx%5E3%7D%7B3%7D%7D%5D%5E2__0)
![R = 3.185 [2 + \frac {2^3}{3}}]](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5B2%20%2B%20%5Cfrac%20%7B2%5E3%7D%7B3%7D%7D%5D)
R = 14.863 Ω
Since V = IR


I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
A. The object is made of stone or metal, which always sink.
B. The force of gravity is stronger than the buoyant force.
C. The buoyant force is stronger than the force of gravity.
D. The mass of the item is greater than its gravity
The answer is C