Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J
Answer:
Torque τ =w ×0 = 0
Explanation:
We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.
Here the gravity force act at the center and the rotational axis is also passing through the center.
Therefore the perpendicular distance between the force and the rotational axis would be zero.
Hence the torque will be
Torque = Force × Perpendicular distance
Torque = mg×0 = 0
Therefore the torque would be zero.
Answer: 6.48m/s
Explanation:
First, we know that Impulse = change in momentum
Initial velocity, u = 19.8m/s
Let,
Velocity after first collision = x m/s
Velocity after second collision = y m/s
Also, we know that
Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).
5700 = 1500(19.8 - x)
5700 = 29700 - 1500x
1500x = 29700 - 5700
1500x = 24000
x = 24000/1500
x = 16m/s
Also, at the second guard rail. impulse = ft, so that
Impulse = 79000 * 0.12
Impulse = 9480
This makes us have
Impulse = m(x - y)
9480 = 1500(16 -y)
9480 = 24000 - 1500y
1500y = 24000 - 9480
1500y = 14520
y = 14520 / 1500
y = 9.68
Then, the velocity decreases by 3.2, so that the final velocity of the car is
9.68 - 3.2 = 6.48m/s
Answer:
Impulse = 90
Resulting Velocity = 89
Explanation:
Use F * change in time = m * change in velocity.
For the first part of the question, the left side of the equation is the impulse. Plug it in.
60 * (3.0 - 0) = 90.
For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.
60 * (3.0 - 0) = 10 * (vf - 80). Simplify.
90 = 10vf - 800. Simplify again.
890 = 10vf. Divide to simplify and get the answer.
The resulting velocity is 89.
