Answer:
17.6kg.m/s in the opposite direction of the initial velocity.
Explanation:
In order to solve this we can use the Impulse-Momentum theorem:
We will assume the motion of the ball approaching the player as positive, so:
So the impulse is 17.6kg.m/s in the opposite direction of the initial velocity.
1.244 m per second the person driving will go
Answer: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is
where
is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m
