Answer:
Explanation:
Given:
Initial velocity of the vehicle, 
distance between the car and the tree, 
time taken to respond to the situation, 
acceleration of the car after braking, 
Using equation of motion:
..............(1)
where:
final velocity of the car when it hits the tree
initial velocity of the car when the tree falls
acceleration after the brakes are applied
distance between the tree and the car after the brakes are applied.
Now for this situation the eq. (1) becomes:
(negative sign is for the deceleration after the brake is applied to the car.)
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω
Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂
The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁
Hence, the protostar will be rotating 16 </span><span>times faster.</span>
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.
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
