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Katen [24]
3 years ago
7

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height

in feet after t seconds is given by y = 27 t − 20 t 2 y=27t-20t2. A. Find the average velocity for the time period beginning when t=3 and lasting .01 s: Preview .005 s: Preview .002 s: Preview .001 s: Preview NOTE: For the above answers, you may have to enter 6 or 7 significant digits if you are using a calculator. Estimate the instanteneous velocity when t=3. Preview Get help: Video
Physics
1 answer:
Ahat [919]3 years ago
5 0

Answer:

V_{3.01}=-93.2m/s

V_{3.005}=-93.1m/s

V_{3.002}=-93.04m/s

V_{3.001}=-93.02m/s

V_{3}=-93m/s

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

Y_{3}=-99m/s

Y_{3.01}=-99.932m/s

Y_{3.005}=-99.4655m/s

Y_{3.002}=-99.18608m/s

Y_{3.001}=-99.09302m/s

Replacing these values into the formula for average velocity:

V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s

V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s

V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s

V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s

To know the actual velocity, we derive the position and we get:

V=27-40t = -93m/s

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In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t
muminat
<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So,  m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is  m/s2

Acceleration  m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation  ,

m/s

Therefore, the final velocity or jumping speed is  m/s

Explanation:

3 0
3 years ago
Read 2 more answers
A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a
Kryger [21]

Answer:

2583.9 N/C

Explanation:

Parameters given:

Outer diameter = 14 cm

Outer radius, R = 7cm = 0.07m

Inner diameter = 7 cm

Inner radius, r = 3.5 cm = 0.035m

Charge of washer = 8 nC = 8 * 10^(-9)C

Distance from washer, z = 33 cm = 0.33m

The electric field due to a washer (hollow disk) is given as:

E = k * σ * 2π [ 1 - z/(√(z² + R²)]

Where σ = charge per unit area

σ = q/π(R² - r²)

σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)

σ = 2.077 * 10^(-6) C/m²

=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]

E = 117.467 * 10^3 * (1 - 0.978)

E = 117.467 * 10^3 * 0.022

E = 2583.9 N/C

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3 years ago
Think about a car design. How would you improve efficiency? What new features would your car be able to do? Write at least 5 dif
drek231 [11]

Answer:

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big

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Explanation:

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Why are the orbits of planets only nearly circular and not perfectly circular?
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Explanation:

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2 years ago
You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is
madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book(m_0)=1.7 kg

height(h)=1 m

time taken=0.64 s

h=ut+frac{at^2}{2}

1=0+\frac{a(0.64)^2}{2}

a=4.88 m/s^2and [tex]a=\alpha r

where \alphais angular acceleration of pulley

4.88=\alpha \times 5.2\times 10^{-2}

\alpha =93.84 rad/s^2

And Tension in Rope

T=m(g-a)

T=1.7\times (9.8-4.88)

T=8.364 N

and Tension will provide Torque

T\times r=I\cdot \alpha

8.364\times 5.2\times 10^{-2}=I\times 93.84

I=0.463\times 10^{-2} kg-m^2

I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2

Thus mass is uniformly distributed or some more towards periphery of Pulley

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