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3 years ago

1. A garelle running 14 m/s does not see the high cliff ahead. The next day buzzards are seen cireling a

Physics

1 answer:

katovenus [111]3 years ago

8 0

The vertical motion of a dropped object from height is a motion that is caused by gravity alone

The height of the cliff and the speed of the bullet are:

1. The height of the cliff is 19.6 meters

2. The speed of the bullet when it left the gun is approximately 2,245 m/s

The reason the above values are the correct values are given as follows:

The given velocity of the gazelle, vₓ = 14 m/s

Location where buzzards are seen, d = 28 m from the base of the cliff

The height of the cliff = Required

Solution:

The gazelle fell off the cliff

The horizontal distance the gazelle covered during free fall, d = 28 meters

The time it takes the gazelle to travel the horizontal distance, t = The time of free fall, and is given as follows;

$t = \dfrac{Horizontal \ distance}{Horizontal \ velocity} = \dfrac{d}{v_x}$

Therefore;

$t = \dfrac{28 \ m}{14 \ m/s} = 2 \ s$

Vertical distance covered in the time falling freely = Height of the cliff, h

The formula for (vertical) distance, h = u·t + (1/2)·g·t²

Where;

u = The initial vertical velocity = 0

t = 2 seconds from above

g = Constant for the acceleration due to gravity ≈ 9.8 m/s²

Therefore;

h = 0 × 2 + (1/2) × 9.8 m/s² × (2 s)² ≈ 19.6 m

The height of the cliff, h = 19.6 m

2. The known parameters:

The direction the bullet was shot = Horizontally

Height from which the bullet was shot, h = 1.4 meters above the ground

The horizontal distance the bullet travels before landing, d = 1.2 km

Required:

The speed with which the bullet left the gun, assuming no air resistance

Solution:

The vertical motion of the bullet is due to gravity and the initial vertical velocity, u = 0

Therefore;

h = u·t + (1/2)·g·t²

1.4 = (1/2) × 9.8 × t²

t² = 1.4/((1/2) × 9.8) = 2/7

t = √(2/7)

$Velocity = \dfrac{Distance}{Time}$

$The\ speed \ of \ the \ bullet, v = \dfrac{1,200 \ m}{\sqrt{\dfrac{2}{7} } s} \approx 2,245 \ m/s$

The speed of the bullet when it left the gun, v ≈ 2,245 m/s

Learn more about free fall motion here:

brainly.com/question/2148964

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Answer:

The temperature is $T = 168.44 \ K$

Explanation:

From the question ewe are told that

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Generally, the rate of heat transfer is mathematically represented as

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=> $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where $\sigma$ is the Boltzmann constant with value $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value

$T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

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3 years ago

A sample contains 36 g of a radioactive isotope. How much radioactive isotope remains in the sample after 3 half-lives?

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Answer:

Option "C": "4.5 g"

Explanation:

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t = 3 T, n is number of half lives = t / T = 3

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N / N0 = (1 / 2)^n

where

"N0" be the "initial amount"

"N" be the "amount left"

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3 years ago

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What is Secular Music's instrument's?

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Explanation:

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3 years ago

A person weighing 645 N climbs up a ladder to a height of 4.55 m what is the increase in gravitational energy

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Answer: 2934.75 Joules

Explanation:

Potential energy can be defined as energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

P.E = mgh

Where P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per second square.

h represents the height measured in meters.

Given the following data;

Weight =645

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But we know that weight = mg = 645N

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P.E = 645 • 4.55

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Potential energy, P.E = 2934.75 Joules.

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3 years ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

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3 years ago