Part e a small toy cart equipped with a spring bumper rolls toward a wall with a speed of v. the cart rebounds from the wall, wi
th the same speed v. the sketch below shows the initial velocity vector v⃗ i, final velocity vector v⃗ f, and the initial momentum vector p⃗ i. using the initial momentum vector as a basis, draw the change in momentum vector δp⃗ for the cart. you can obtain each vector's length and orientation by clicking on it to select it.
1 answer:
Answer:
Δp = -2 p₀
Explanation:
Momentum is defined as
p = m v
In this instance, we denote the initial and final momentum, with positive direction considered towards the wall.
p₀ = m v
p_f = m (-v)
The negative sign indicates the car is rebounding off the wall.
The change in momentum is expressed as
Δp = p_f - p₀
Δp = - m v - m v
Δp = -2 mv
Δp = -2 p₀
This shows that the change in momentum equals twice the initial momentum; the vectors demonstrating these changes can be seen in the attachment, and the sign signifies the direction of this change.
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Response:
Clarification:
Refer to the diagram indicating the charges on the specified sphere (see attachment).
The electric field at the stated positions is
E(r) = 0 for r≤a. Equation 1
E(r) = kq/r² for a<r<b. Equation 2
E(r) = 0 for b<r<c. Equation 3
E(r) = kq/r² for r>c. Equation 4.
We understand that electric potential correlates with the electric field through
V = Ed
A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is
E = kQ / r²
Then,
V = Ed,
At d = r = c
Thus,
Vc = (kQ / c²) × c
Vc = kQ / c
As a result, the total charge Q consists of +q, -q, and +q
Hence, Q = q - q + q = q
V = kq / c
B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have
V = kQ/r
V = kQ / b, noting that r = b
So, Q = q
V = kq / b
C. At r = a
Following from equation 1:
E(r) = 0 for r≤a. Equation 1
The electric field at the surface of the solid sphere is 0, E = 0N/C
Thus,
V = Ed = 0 V
Consequently, the electric potential at the solid sphere's surface is 0.
D. At r = 0
The electric potential can be determined by
V = kq / r
As r approaches 0,
V = kq / 0
V approaches infinity.
Answer: Tension = 47.8N, Δx = 11.5× m.
Tension = 95.6N, Δx = 15.4× m
Explanation: The speed of a wave on a string under tension can be determined using the following:
denotes tension (N)
μ refers to linear density (kg/m)
Calculating the velocity:
0.0935 m/s
Distance a pulse traveled in 1.23ms:
Δx = 11.5×
With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5× m.
When tension is doubled:
|v| = 0.1252 m/s
Distance in the same time:
15.4×
With the increased tension, it moves 15.4×
m