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zloy xaker [14]
3 years ago
8

Traveling with an initial speed of a car accelerates at along a straight road. How long will it take to reach a speed of Also, t

hrough what distance does the car travel during this time? (10%) b. At bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur
Physics
1 answer:
makvit [3.9K]3 years ago
6 0

Answer:

A) 30 s, 792 m

B) 10.28 s, 4108.2 m = 4.11 km

Explanation:

A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

Using the equations of motion.

v = u + at

v = final velocity = 120 km/h

u = initial velocity = 70 km/h

a = acceleration = 6000 km/h²

t = ?

120 = 70 + 6000t

6000t = 50

t = (50/6000) = 0.0083333333 hours = 30 seconds.

Using the equations of motion further,

v² = u² + 2ax

where x = horizontal distance covered by the car during this time

120² = 70² + 2×6000×x

12000x = 120² - 70² = 9500

x = (9500/12000) = 0.79167 km = 791.67 m = 792 m

B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

For the bullet A

u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

t = T

y = 600T - 4.9T²

At the point where the two bullets pass each other, the vertical heights covered are equal

y = y

450t - 4.9t² = 600T - 4.9T²

But, note that, since T starts reading, 3 seconds after t started reading,

T = (t - 3) s

450t - 4.9t² = 600T - 4.9T²

450t - 4.9t² = 600(t-3) - 4.9(t-3)²

450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)

450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1

600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0

179.4t - 1844.1 = 0

t = (1844.1/179.4) = 10.28 s

Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.

y = 450t - 4.9t²

= (450×10.28) - (4.9×10.28×10.28)

= 4,108.2 m = 4.11 km

Hope this Helps!!!!

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Answer:

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Explanation:

From the question we are told that

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Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

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Thus  

       ma  =  \frac{mv^2}{r}

=>    \frac{v^2}{r}  =  a

      Generally the speed of this spaceship is mathematically represented as

      v =  \frac{2 \pi}{T}

=>    v^2  =   [\frac{2\pi}{T}] ^2

=>     \frac{\frac{4\pi^2 r^2}{T^2} }{r}  = 0.5g

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<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

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The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\  \frac{m_1}{m_2\times 4} = 1 \\ \\  \frac{m_1}{m_2} = 4

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

 

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2 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
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Answer:

The extension of the wire is 0.362 mm.

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length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

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The downward force of the object on the wire is given by;

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The Young's modulus of aluminum is given by;

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Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

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Answer:

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Explanation:

QUESTION 1:

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Where;

v = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration due to gravity (m/s²)

s = distance (m)

From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²

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QUESTION 2:

Using v = u + at

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a = acceleration due to gravity (m/s²)

v = u + at

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20.54 - 5.65 = 9.8t

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t = 14.89/9.8

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t = 1.52s

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Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

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