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2 years ago

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A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d

egrees south of that direction, and to continue to drive 50 miles deviating by 30 degrees each time until he returns to his original position. How far will he drive, and how many vectors must he sum to calculate his displacement

1 answer:

stira [4]2 years ago

5 0

Answer:600 miles, 12

Explanation: The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.

The sum of exterior angles of a polygon = 360 degrees.

Exterior angle of a polygon = (360 ÷ number of sides)

Therefore,

Number of sides = 360 ÷ exterior angle

Exterior angle = 30 degrees

Hence,

Number of sides = 360 ÷ 30 = 12 sides

Since distance traveled of 50 miles is the same for each displacement ;

Total displacement = distance traveled * number of sides

Total displacement = 50 * 12 = 600 miles.

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A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet an

kykrilka [37]

Answer:

a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

$m_Hv_H + m_Av_A = m_Hv_2+m_Av_2$

From the question given, we can deduce that the anvil is at rest;

∴ $v_A = 0$ ; then, we have:

$m_Hv_H + 0 = (m_H+m_A) v_2$

Making $v_2$ the subject of the formula; we have:

$v_2 = \frac{m_Hv_H}{m_H + m_A}$ ------- Equation (1)

Also, from the second diagram; there is a representation of a free body momentum of the hammer head;

From the diagram;

F = impulsive force exerted on the rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

$m_Hv_H - F \delta t = m_Hv_2$

$- F \delta t = - m_Hv_H + m_Hv_2$

$F \delta t = m_Hv_H - m_Hv_2$

$F \delta t = m_H(v_H - v_2)$ ------- Equation (2)

Using the function of the kinetic energy of the hammer before impact $T_1$ ; we have:

$T_1 = \frac{1}{2} m_Hv_H^2$ -------- Equation (3)

We determine the mass of the hammer $m_H$ by using the formula from:

$W_H = m_Hg$

where;

$W_H$ = weight of the hammer

$m_H$ = mass of the hammer

g = acceleration due to gravity

Making $m_H$ the subject of the formula; we have:

$m_H = \frac{W_H}{g}$

$m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}$

$m_H = 0.04658 \ lb.s^2/ft$

Now;

$T_1 = \frac{1}{2} m_Hv_H^2$

$T_1 = \frac{1}{2}*(0.04658 \ lb.s^2 /ft) *(20 \ ft/s)^2$

$T_1 = \frac{18.632 }{2}$

$T_1 = 9.316 \ ft.lb$

After the impact $T_2$ ; the final kinetic energy of the hammer and anvil can be written as:

$T_2 = \frac{1}{2}(m_H +m_A)v^2_2$

Recall from equation (1) ; where $v_2 = (\frac{m_Hv_H}{m_H+m_A})$ ; if we slot that into the above equation; we have:

$T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2$

$T_2 = \frac{1}{2} \frac{m^2_H +v^2}{m_H+m_A}$

$T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})$

Also; from equation (3)

$T_1 = \frac{1}{2} m_Hv_H^2$ ; Therefore;

$T_2 = T_1 (\frac{m_H}{m_H+m_A})$ ----- Equation (4)

a)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil $m_A = \frac{W_A}{g}$

where we replace; $W_A \ with \ \infty$ and g = 32.2 ft/s²

$m_A = \frac{\infty}{32.2 \ ft/s}$

However ; from equation (1)

$v_2 = \frac{m_H v_H}{m_H + m_A}$

$v_2 = \frac{0.04658*20}{0.04658+ \ \infty}$

$v_2 = 0$

From equation (2)

$F \delta t = m_H(v_H + v_2)$

$F \delta t = (0.04658 lb .s^2 /ft )(20ft/s - 0)$

$F \delta t = \ 0.932 \ lb.s$

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

For the energy absorbed by the rivet ; we have:

$T_2 = T_1 (\frac{m_H}{m_H+m_A} )$

where;

$T_1= 9.316 \ ft.lb$

$m_H = 0.04658 \ lb.s^2/ft$

$m_A = \infty$

Then;

$T_2 = (9.316 \ ft.lb) (\frac{0.04658\ lb.s^2/ft)}{0.04658 \ lb.s^2/ft+ \infty} )$

$T_2 = (9.316 \ ft.lb)* 0$

$T_2 = 0$

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = $T_1 - T_2$

ΔT = 9.316 ft.lb - 0

ΔT ≅ 9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b)

Due to the broadness of this question, the text is more than 5000 characters, so i was unable to submit it after typing it . In the bid to curb that ; i create a document for the answer for the part b of this question.

The attached file can be found below.

5 0

2 years ago

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago

What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

densk [106]

Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c

7 0

2 years ago

Read 2 more answers

A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting

Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

$\Delta U = n C_v(T_f-T_i)$

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

$\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K$

So, $\Delta U= 0.2\times12.5(1200-300)\\=2250 J$

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

$\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J$

Since, Q is positive 2250 J of heat will flow into the system.

6 0

2 years ago

Select the areas that would receive snowfall because of the lake effect.

Studentka2010 [4]

Answer:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The places that would would get snowfall because of the lake effect are Grand Marais, Two Harbors, and Duluth. The reason for this is that these three places are located right on the shores of the Lake Superior. This lake is one of the biggest lakes in the world. It has enormous amount of water in it, having big impact on the regional climate because of that. The water from this lake creates a lot of humidity in the air, and there's a lot of evaporation as well, both causing the formation of clouds, and when it is cold enough, instead of precipitation, the region gets large amounts of snowfall.

8 0

2 years ago