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Sonja [21]
3 years ago
8

A 200 g air-track glider is attached to a spring. The glider is pushed 10.0 cm against the spring, then released. A student with

a stopwatch finds that 10 oscillating take 12.0 s. What is the spring constant?
Physics
1 answer:
fiasKO [112]3 years ago
4 0

Answer:

5.5N/m

Explanation:

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

k=4π²m/T²

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

k=4π²×0.2kg/(1.2)²

k=39.48×0.2kg/1.44

k=7.90/1.44

k=5.48N/m

k=5.5N/m ( Approximately)

Therefore the spring constant will be 5.5N/m

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Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

where:

q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

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W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

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3 years ago
An electrical power plant manages to send 88% of the heat produced in the burning of fossil fuel into the water-to-steam convers
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Answer:

A

Explanation:

Let the x represent the amount of heat generated from the fossil fuel.

88% of x = 0.88 x

0.88 x was used to convert water to steam.

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efficiency of the heat -to- work conversion = work output / work input = 0.352 x / x = 0.352 × 100 = 35.2 % which is less than 40 %

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3 years ago
A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
Crazy boy [7]

Answer:

Temperature at the exit = 267.3 C

Explanation:

For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

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