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3 years ago

An arrow is shot at 28.0° above the horizontal. Its initial speed is 50 m/s and it hits the target.

1 answer:

fomenos3 years ago

6 0

We will need to work with the components of the velocity, in the x and the y direction. We will say up is positive so g is -9.81 m/s^2.

Given that the angle was 32 degrees:

Velocity up (in the y direction) is 55 m/s * sin 32 = 29.15 m/s
And

Velocity forward (in the x direction) is 55 m/s * cos 32 = 46.64 m/s

The acceleration of gravity, -9.81 m/s2 continuously decreases the velocity in the y direction. At the maximum height, the velocity will be zero. This should make sense, for as soon as the decreasing velocity becomes negative, the arrow will start to fall.

We have v = v(0) + at

And we set this to zero and solve for t:

0 = 29.15 + -9.81t

9.81t = 29.15

t = 2.97 seconds

To calculate height at this point, we use the equation that calculates position based on time, acceleration, and initial velocity (we could use an alternate too, an equation derived from the one we are now using and v = v(0) + at.

x = x(0) + v(0)t + (1/2)at^2

x = 0 + 29.15 * 2.97 + 0.5 9.81 (2.97)^2

x = 43.30 m

For a projectile, the plot of distance traveled in the upward direction is a parabola, and it takes the same amount of time to come down as it did to go up.

We can double 2.97 to get the time of impact on the target at 2(2.97) = 5.94 seconds

(Alternately, if you like, you can solve

0 = 0 + 29.15t + 0.5 9.81 t^2

And find that the two roots are 0 and 5.94).

http://www.math.com/students/calculators... will do the quadratic for you.

Given a horizontal velocity of 46.64 m/s, we can calculate

46.64 m/s (5.94 s) = 277 m for the distance of the target.

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Find equailent resistance

vovikov84 [41]

Answer:

0.71ohms

Explanation:

AC and CB are in series hence the total resistance is 3+2=5ohms

AD and DB are in series hence the total resistance is 3+2=5ohms

These resistances are parallel to resistance AB

Let the equivalent resistance be R.

For resistances in parallel

Hence the total resistances are:

5//1//5

1/R =1/5 + 1/1 +1/5 = 1+5+1 /5 = 7/5

Hence the total resistance is R= 1/ 7/5 = 1×5/7= 5/7=0.71ohms

4 0

3 years ago

An object of mass 30kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.

Luba_88 [7]

That all depends on the planet toward which the mass is falling.

If this happens to be taking place near the Earth, then the object accelerates
at the rate of about 9.8 meters per second every second.

Furthermore, if there is truly no air resistance, then it makes no difference whether
the object is a feather, a mass of 30 kg, or a school-bus. All objects accelerate at
the same rate regardless of their mass.

7 0

3 years ago

A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After

Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

$v^2 -u^2 = 2ad$

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

$v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s$

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

$d=\frac{1}{2}at^2$

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s$

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

$K_i = U_f\\\frac{1}{2}mu^2 = mgh$

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

$h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m$

So the maximum altitude reached by the rocket is

$h' = d+h=1500 m +550.8 m=2050.8 m$

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

$h=\frac{1}{2}gt^2$

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s$

So the total time of the trip is

$t'=28.9 s+10.6 s=39.5 s$

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

$\frac{1}{2}mv^2 = mgh$