Answer:
M V = m v conservation of momentum (Caps-cannon Small-projectile)
V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon
KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J
Answer:
The magnitude of vector B is 43 units and it points in the negative y-direction.
Explanation:
Resultant of vectors = vector sum of all the vectors
Vector A = 29j
Vector B = ?
Resultant of vector A and B = R = -14j
R = A + B
-14j = 29j + B
B = -14j - 29j = - 43j
Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.
The answer is:
C. 361 m/s
The explanation:
To calculate the speed of sound at a given temperature (50°C) we are going to use this formula:
v = 331 + 0.6T
when V is the velocity
and T is the temperature = 50°C
by substitution:
v = 331 + 0.6(50)
v = 361 m/s
So, The correct answer is C.
because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the sound in the air is change with temperature.
Answer:
F₃ = 122.88 N
θ₃ = 20.63°
Explanation:
First we find the components of F₁:
For x-component:
F₁ₓ = F₁ Cos θ₁
F₁ₓ = (50 N) Cos 60°
F₁ₓ = 25 N
For y-component:
F₁y = F₁ Sin θ₁
F₁y = (50 N) Sin 60°
F₁y = 43.3 N
Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:
F₂ₓ = F₂ = 90 N
F₂y = 0 N
Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:
F₁ₓ + F₂ₓ + F₃ₓ = 0 N
25 N + 90 N + F₃ₓ = 0 N
F₃ₓ = - 115 N
for y-components:
F₁y + F₂y + F₃y = 0 N
43.3 N + 0 N + F₃y = 0 N
F₃y = - 43.3 N
Now, the magnitude of F₃ can be found as:
F₃ = √F₃ₓ² + F₃y²
F₃ = √[(- 115 N)² + (- 43.3 N)²]
<u>F₃ = 122.88 N</u>
and the direction is given as:
θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)
<u>θ₃ = 20.63°</u>
When the car moves and makes a sound that is louder that when the car is just sitting there
