Ask question

Ask question

All categories

3 years ago

6

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho

rizontal. It explodes on the mountainside 36.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?

1 answer:

cupoosta [38]3 years ago

5 0

Answer:

xf = 5.68 × 10³ m

yf = 8.57 × 10³ m

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0 yi = 0

xf = ? yf = ?

vxi = vicosθ vyi = visinθ

ax = 0 ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t² ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ² ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²

yf = 8.57 × 10³ m

You might be interested in

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu

romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0

2 years ago

Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig

MissTica

Answer:

v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

Em_f = K = ½ m v_f²

Emo = Em_g

½ m v₀² + m g y₀ = ½ m v_f²

v_f² = v₀² + 2 g y₀

let's calculate

v_f = √(10² + 2 9.8 25)

v_f = 24.3 m / s

6 0

2 years ago

Compare and contrast the particle motion in each state of matter

Vladimir79 [104]

Answer:

Gases, liquids and solids are all made up of microscopic particles, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

<h3>Hope this is fine for you</h3>

6 0

3 years ago

Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p

irina1246 [14]

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = $\frac{4}{2}$ =2ohm [ proven as in series $R_{e} = 2 + 2 = 4ohm$ ]

to calculate the equivalent resistance when in parallel:

$\frac{1}{R_{e} } = \frac{1}{R_{1} } + \frac{1}{R_{2}}$

so,

$\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}$

$R_{e} = 1ohm$

4 0

2 years ago