The ____ force is the sum of all the forces that act upon each other

andrew-mc [135]

To convert parametric to Cartesian systems, you need to find a way to get rid of the t's.

In this case, the t's are inside trigonometric functions, so we're going to use a very famous trig identity you should memorize:

${sin(t)}^{2} + {cos(t)}^{2} = 1$

If we plug sin(t) and cos(t) into that equation only x and y variables will be left!

BUT there's one thing. The given cos(t + pi/6) has nasty extra stuff in it. However, part a gives you a tip on how to relate x and y to a nice clean cos(t)

So if we do a little rearranging:

$\sin(t) = \frac{y}{2} \\ \cos(t) = \frac{x + y}{2 \sqrt{3} }$

Now we can plug these into the famous trig identity!

${( \frac{y}{2}) }^{2} + {( \frac{x + y}{2 \sqrt{3} } )}^{2} = 1$

Do a little bit of adjustments to get that final form asked for, and you'll be able to find those integers of a and b. ;)

7 0

3 years ago

In the image, the net force is _______.

Sveta_85 [38]

Your correct answer would be 0 N

The forces are pulling in opposite directions, therefore they cancel each other out.

Hope that helps!!

:)

4 0

3 years ago

The value of k is given as 9.0 x 109 N.m^2/C^2. Which unit must be used

Zinaida [17]

The answer for this question is: C

6 0

3 years ago

Read 2 more answers

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration ModifyingAbove a With right-arrow e

Mandarinka [93]

To answer this problem we will make use of two of the equations of motion. Then we will come across out each velocity component. The difficult part comes in solving for the j component of velocity for which we will have to know t_f, the time it takes to be displaced (11m):

Our equations
1. v_f^2 = v_0^2 + 2ax

2. v_f = v_0 + at

Solving for the time it takes to travel 11m

We do so by finding the final velocity using (1), and then plugging that back into (2).

v_fi^2 = 4^2 + 2*5*11

v_fi^2 = 16 + 110 = 126

v_fi = 11.22 m/s ((()))

plugging this into (2)

11.22 = 4 + 5*t_f

5*t _f= 7.22

t_f = 7.22 / 5

t_f = 1.444 seconds

Solving for the vertical velocity

v_fj = v_0j + at

v_fj = 0 + 7*1.444

v_fj = 10.108 m/s ((()))

Finding magnitude and angle

V = sqrt(v_fi^2 + v_fj^2)

125.884 + 10.108

V = 15.10151 m/s

Angle:

Theta = arctan(v_fj/v_fi)

Theta = 0.0801 radians

converting radians to degrees (180/pi)

Theta = .0.0801(180/pi) deg= 4.591 deg

Answer:

V = 15.10151 m/s

Theta = 4.591 deg

6 0

3 years ago

A projectile (mass = 0.15 kg) is fired at and embeds itself in a stationary target (mass = 2.44 kg). With what percentage of the

Bess [88]

Answer:

38.6 %

Explanation:

First of all, we have to calculate the final velocity of the block-bullet system. We can apply the law of conservation of momentum:

$m u + M U = (m+M)v$

where

m = 0.15 kg is the mass of the bullet

u is the initial velocity of the bullet

M = 2.44 kg is the mass of the block

U = 0 is the initial velocity of the block (it is at rest)

v is the final velocity of the bullet+block

Solving for v,

$v=\frac{mu}{m+M}$

The total initial kinetic energy of the system is just the kinetic energy of the bullet:

$K = \frac{1}{2}mu^2$

While the final kinetic energy of the block+bullet is:

$K' = \frac{1}{2}(m+M) v^2 = \frac{1}{2}(m+M) \frac{(mu)^2}{(m+M)^2}=\frac{1}{2} \frac{m}{m+M}u^2$

So the fraction of kinetic energy lost is

$\frac{K-K'}{K}=\frac{\frac{1}{2}mu^2 - \frac{1}{2}\frac{m}{m+M}u^2}{\frac{1}{2}mu^2}=\frac{m-\frac{m}{m+M}}{m}=\frac{0.15-\frac{0.15}{0.15+2.44}}{0.15}=0.614$

And so, the fraction of kinetic energy left in the projectile after he flies off the block is

1 - 0.614 = 0.386 = 38.6 %

4 0

3 years ago