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Troyanec [42]
3 years ago
10

A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio

n of the particle at the instant when its velocity is zero
Physics
1 answer:
8090 [49]3 years ago
4 0

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

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kakasveta [241]

Answer: 0.5 seconds or 2.625 seconds

Explanation:

At t = 0, The ball is 4 ft above the ground.

The height of the football varies with time in the following way:

s(t) = -16 t² + 50 t + 4

we need to find the time in which the height would of the football would be 25 ft:

⇒25 = -16 t² + 50 t + 4

we need to solve the quadratic equation:

⇒ 16 t² - 50 t + 21 = 0

t = \frac{50 \pm \sqrt{50^2-4\times 16\times 21}}{2\times16}

⇒ t = 0.5 s or 2.625 s

Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.

3 0
3 years ago
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A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = 0.355m

This means that the building will be 35.5cm taller

3 0
3 years ago
A car traveling with constant speed travels 150 km in 7200 s what is speed of the car
spayn [35]

The speed of the car is exactly 150/7200 km/sec, or 125/6 meters/sec. 

In more familiar units, that speed is equivalent to ...

-- (20 and 5/6) meters/sec

-- 75 km/hour

8 0
3 years ago
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3 years ago
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The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:
Furkat [3]
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is  s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and w1 = 1 revolution/sec = 2Pi (rad/s)
Now we can go to proportion
v1=v2
w1*r1 = w2r2w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)
w2 = w3 (which is the   angular velocity of the rear wheel) &#10;
SOLVING FOR A : v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec
v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)= 9.99 or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!
7 0
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