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2 years ago

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

Physics

1 answer:

tamaranim1 [39]2 years ago

4 0

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

$s=ut+\frac{1}{2}*gt^{2} \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m$

Arrow misses the target by 5.843m ig the arrow us split horizontally

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You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅ = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0

1 year ago

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0

2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

2 years ago

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to fi

JulijaS [17]

Answer:

b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

w₀² = k / m

w² = w₀² - b² / 4m²

b² = (w₀² -w²) 4 m²

Let's find the angular velocities

w₀² = 5 / 0.1

w₀² = 50

w = 2π f

w = 2π 1

w = 6.2832 rad / s

we subtitute

b² = (50 - 6.2832²) 4 0.1²

b = √ 0.42086

b = 0.6487 kg / s

8 0

2 years ago

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

3 0

2 years ago