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3 years ago

Calculate the speed for each segment on the graph include the units (m/s). Show your calculations, (such as 4/2= 2 m/s)

Physics

1 answer:

fgiga [73]3 years ago

6 0

Answer:

v_A = 6 m/s , v_B = 0 m/s , v_C = - 4 m/s , v_D = -4 m/s , v_E = 2.66 m / s

Explanation:

For this exercise we use the definition of average velocity in each segment

v_average = (x₂-x₁) / Δt

segment A of the graph you take the value of distance and time

x₁ = 0 m t₁ = 0 s

x₂ = 60 m t₂ = 10 s

using the average velocity equation and calculate

v_A = (60-0) / (10-0)

v_A = 6 m / s

Segment B

x₁ = 60m

x₂ = 60 m

since the displacement variation is zero (particle stopped) the velocity is zero

v_B = 0 m / s

Segment C and D

x₁ = 60 m t1 = 15 s

x₂ = -40 m t2 = 40 s

v_C = (-40 -60) / (40 -15)

v_C = - 4 m / s

Segment D

how is the same line

v_D = -4 m / s

Segment E

x₁ = -40 m t₁ = 40 s

x₂ = 0 m t₂ = 55 s

v_E = (0- (-40)) / (55 -40)

v_E = 2.66 m / s

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A vacuum gage connected to a chamber reads 39 kPa at a location where the atmospheric pressure is 92 kPa. Determine the absolute

Nata [24]

Answer: 53 kPa

Explanation:

Absolute pressure is a pressure value referred to absolute zero or vacuum. This value indicates the total pressure to which a body or system (the chamber in this situation) is subjected, considering the total pressure acting on it.

In this sense, the equation that will be useful in this case is:

$P_{atmospheric}= P_{absolute} + P_{vacuum}$ (1)

Where:

$P_{atmospheric}=92 kPa$ is the atmospheric pressure

$P_{vacuum}=39 kPa$ is the vacuum pressure

$P_{absolute}$ is the absolute pressure

Isolating $P_{absolute}$ from (1):

$P_{absolute}=P_{atmospheric} - P_{vacuum}$ (2)

$P_{absolute}=92 kPa - 39 kPa$ (3)

Finally:

$P_{absolute}=53 kPa=53(10)^{3} Pa$ This is the absolute pressure in the chamber

7 0

4 years ago

The 15 g head of a bobble-head doll oscillates in SHM at a frequency of 4.0 Hz.

Goshia [24]

Answer:

(a) 9.375 N/m

(b) 0.5024 m/s

Explanation:

mass of head, m = 15 g = 0.015 kg

frequency, f = 4 Hz

Time period, T = 1 / f = 0.25 s

Let k is the spring constant.

(a)

The formula for the time period is

$T=2\pi\sqrt{\frac{m}{K}}$

$0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}$

$0.04=\sqrt{\frac{0.015}{K}}$

K = 9.375 N/m

(b)

Amplitude, A = 2 cm

Let ω is the angular velocity.

Maximum velocity, v = A ω = A x 2πf

v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s

(c)

Let b is the damping constant.

A(t = 4s) = 0.5 cm

Ao = 2 cm

Using the formula of damping

$\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}$

$\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}$

$0.25=e^{-133.3 b}$

Taking natural log on both the sides

ln (0.25) = - 133.3 b

- 133.3 b = - 1.386

b = 0.01 kg/s

3 0

3 years ago

A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before co

Tema [17]

<span>This problem is solved by the equation of motion: x = x0 + v0*t + 1/2*a*t^2, Here x0 = 0, v0 = 40ft/sec and a = -5 ft/s^2, we need to solve for t: v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t --> a*t = -v0 --> t = -v0/a -- > 40/5 = 8 seconds to stop. In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft. Answer: The car travels 160 ft.</span>

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3 years ago

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Svet_ta [14]

Answer:

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Explanation:

See the picture attached