Equations To Remember For Ch01.04.05.06 Test

A. (-0.620,109°)

1.A {Standard Form} Graph, as line-segments, the following polar coordinates on Cartesian Coordinate System:

Line segments, closed styles

B. (0.276,61.00°)

1.B {Standard Form} Graph, as line-segments, the following polar coordinates on Cartesian Coordinate System:

Line segments, closed styles

Given the equation "x = A + B + C + D + E", where:

A = 0.11m
B = 0.385m
C = 20030m
D = 42600m
E = 51070m

2.1 {Scientific Notation} Solve for [x].

Transcribing for C, D, E it’s ex: 0DPs [P] 1e-01

Final Answer Value ex: 1.370e + 05 (10^5 100000)

Given the equation "x = A + B + C + D + E", where:

A = 0.11m
B = 0.785m
C = 20030m
D = 42600m
E = 51070m

3.1 {Scientific Notation} Solve for [x].

Transcribing for C, D, E it’s ex: 0DPs [P] 1e-01

Final Answer Value ex: 1.370e + 05 (10^5 100000)

As a fish jumps vertically out of the water, assume that only two significant forces act on it (i.e., an upward force [F_up] exerted by the tail fin and the downward force [F_g] due to gravity).  A record Chinook salmon has a length [l] of 1.50m and a mass [m]of 49.5kg.  If this fish is moving upward [v₀] at 3.00m/s as its head first breaks the surface and has an upward speed [v_f] of 6.40m/s after two-thirds of its length has left the surface, determine the following (assuming constant acceleration & g = -9.81m/s²:

4.1 (a) {Standard Form} The salmon's acceleration [a_y].

Solve for Δy first: 2/3 times I → 2/3 times 1.30m

As a fish jumps vertically out of the water, assume that only two significant forces act on it (i.e., an upward force [F_up] exerted by the tail fin and the downward force [F_g] due to gravity).  A record Chinook salmon has a length [l] of 1.50m and a mass [m]of 49.5kg.  If this fish is moving upward [v₀] at 3.00m/s as its head first breaks the surface and has an upward speed [v_f] of 6.40m/s after two-thirds of its length has left the surface, determine the following (assuming constant acceleration & g = -9.81m/s²:

4.2 (b) {Standard Form} The magnitude of the force [F_up] during this interval.

Magnitude of F = N

Two packing crates of masses m₁ = 10.10kg and m₂ = 5.10kg are connected by a light string that passes over a frictionless pulley as in the figure below.    The 5.10kg crate lies on a smooth incline of angle [𝜃] of 41.0°.    Note: g = 9.81m/s².

5.1 (a) {Standard Form} What is the acceleration [a_m2] up the incline of m₂?

Two packing crates of masses m₁ = 10.10kg and m₂ = 5.10kg are connected by a light string that passes over a frictionless pulley as in the figure below.    The 5.10kg crate lies on a smooth incline of angle [𝜃] of 41.0°.    Note: g = 9.81m/s².

5.2 (b) {Standard Form} What is the tension [T] in the string?

Tension = N

Objects with masses m₁ = 12.0kg and m₂ = 7.0kg are connected by a light string that passes over a frictionless pulley as in the figure below.  Note: g = -9.81m/s².

6.1 {Standard Form} If, when the system starts from rest, [m₂] falls a distance [h] of 1.50m in a time [t] of 1.74s, determine the coefficient of kinetic friction [𝜇_k] between [m₁] and the table.

Solve for a: 2h/t²

Solve for N: m1 times |g|

Solve for f: m2 times |g| - (m1+m2) times a

A tension force [T] of 155N inclined at an angle [𝜃] of 30.0° above the horizontal is used to pull a shipping crate, with a mass [m] of 33.0kg, a distance [d] of 5.10m on a rough surface. Note: g = -9.81m/s².

7.1 (a) {Standard Form} If the crate moves at a constant speed, find the work [W_T] done by the tension force [T].

Asking to find the work → J

A tension force [T] of 155N inclined at an angle [𝜃] of 30.0° above the horizontal is used to pull a shipping crate, with a mass [m] of 33.0kg, a distance [d] of 5.10m on a rough surface. Note: g = -9.81m/s².

7.2 (b) {Standard Form} If the crate moves at a constant speed, find the coefficient of kinetic friction [𝜇_k] between the crate and surface.

Hooke's law describes a certain light spring of unstretched length [l_i] of 34.0cm.  When one end is attached to the top of a door frame and an object of mass [m] of 8.00kg is hung from the other end, the length [l_f] of the spring is 44.5cm.  Note: g = -9.81m/s².

8.1 (a) {Standard Form} Find, in kilo Newtons per meter, its spring constant [k].

kilo Newtons per meter → kN/m

Convert cm to m → times .01

Divide N/m by 1000 to get k

Hooke's law describes a certain light spring of unstretched length [l_i] of 34.0cm.  When one end is attached to the top of a door frame and an object of mass [m] of 8.00kg is hung from the other end, the length [l_f] of the spring is 44.5cm.  Note: g = -9.81m/s².

8.2 (b) {Standard Form} The load and the spring are taken down.  Two people pull in opposite directions on the ends of the spring, each with a force [F] of 180.0N.  Find, in centimeters, the length [l_f] of the spring in this situation.

Divide N by 1000 to get kN

Find Δx= F/k

Times Δx by 100cm/m to get cm

Three objects with masses m₁ = 5.40kg, m₂ = 13.0kg, and m₃ = 18.0kg, respectively, are attached by strings over frictionless pulleys as indicated in the figure below.  The horizontal surface exerts a force of friction [f_k] of 31.7N on [m₂].  Note: g = -9.81m/s².

9.1 {Standard Form} If the system is released from rest, use energy concepts to find the speed [v_avg] of m₃ after it moves down [d] 4.10m.

A high-speed photograph of a club hitting a golf ball is shown in the figure below.  The club was in contact with a ball, initially at rest [v₀], for about a time [t] of 0.00320s.

10.1 {Standard Form} If the ball has a mass [m] of 56.5g and leaves the head of the club with a speed [v_avg] of 230.0ft/s, find the average force [F_avg], in kilo Newtons, exerted on the ball by the club.

vavg → vf

top answer of equation times 1kg times 1m times 1N / bottom answer of equation times 10³g times 3.281ft times 1 kgtimesm/s² times 10³N

A ball with a mass [m] of 0.150kg is dropped from rest [v₀] from a height [y_i] of 1.26m.  It rebounds from the floor to reach a height [y_f] of 0.805m. Note:  g = -9.81m/s² and upward it + y-direction

11.1 {Standard Form} What impulse [I] was given to the ball by the floor?

Asking for impulse → Ns

For for v1: -√(voy)²+2ayΔy

For v2: √(vf)²-2ayΔy

I = kg times m/s² times 1N/kg times m/s = Ns

A railroad car with a mass [m_RC] of 2.70 ✕ 10⁴kg moving at a speed [v_avgRC] of 3.15m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at a speed [v_avg2RCs] of 1.26m/s.

12.1 (a) {Standard Form} What is the speed [v_avg3RCs] of the three coupled cars after the collision?

A railroad car with a mass [m_RC] of 2.70 ✕ 10⁴kg moving at a speed [v_avgRC] of 3.15m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at a speed [v_avg2RCs] of 1.26m/s.

12.2 (b) {Standard Form} How much kinetic energy [KE_lost], in kilo Joules, is lost during the collision?

KE lost = kg times m²/s² times 1N/kg times m²/s² times 1J/1Nm times 1kJ/10³J = kJ