Whenever a net force acts on an object, there is a change in the objects ___

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago

Arrange the listed objects according to their angular speeds, from largest to smallest.

Paladinen [302]

Answer:

Explanation:

1 ) tire of radius 0.381 m rotating at 12.2 rpm

12.2 rpm = 12.2 /60 rps

n = .20333 rps

angular speed

= 2πn

= 2 x 3.14 x .20333

= 1.277 rad / s

2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s

angular speed = .456 rad/s

3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second

18.7° per second = (18.7 / 180) x 3.14 rad/s

= .326 rad/s

4 )

a rock on a string being swung in a circle of radius 0.587 m with

a centripetal acceleration of 4.53 m/s2

centripetal acceleration = ω²R

ω is angular velocity and R is radius

4.53 = ω² x .587

ω = 2.78 rad / s

5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s

The radius of the circle in which corner is moving

= .123 x √2

=.174 m

angular velocity = linear velocity / radius

.287 / .174

1.649 rad / s

The perfect order is

4 ) > 5> 1 >2>3.

5 0

3 years ago

If the acceleration of an object is zero at some instant in time, what can be said about its velocity at that time? 1. It is neg

Mkey [24]

3. It is not changing at that time

Explanation:

If the acceleration of a body is zero at some instant in time, it implies that the velocity is not changing at that point in time. Velocity is the rate of change of displacement with time.

✓Acceleration and velocity shares a very close relationship.

✓ For a body to accelerate, the velocity must change. Acceleration is defined as the rate of change of velocity with time.

✓If at any point, a body moves with constant velocity i.e the velocity does not change with time, the acceleration becomes zero.

✓ For acceleration to occur, a body must change velocity.

Learn more:

Acceleration brainly.com/question/6323625

#learnwithBrainly

5 0

4 years ago

In order to distinguish between wavelengths independent of light intensity, one must have at least?

kirill [66]

One must have at least two visual pigments to distinguish between wavelengths independent of light intensity.

A substance that converts light energy into electrical potentials is called a visual pigment.

The basic structure of pigment consists of the chromophore, a colored molecule, and a protein called rhodopsin.

There are three types of visual pigments: Scotopsin, rhodopsins, and photopsin.

Scotopsin pigments are associated with vision and operate in less bright light while photopsin operates in a brighter light.

However, visual pigments are very significant substances as they are associated with vision and play a vital role in detecting light.

If you need to learn more about the wavelength of light click here:

brainly.com/question/9807579

#SPJ4

4 0

2 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

$W = F x = \mu N x = \mu\ m\ g\ x$

$W = 150 \times 6$

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

$\mu = \frac {F}{m\timesg}$

$= \frac{150}{40\times 9.8}$

= .383

We simply applied the above formulas so that each one part could calculate

3 0

4 years ago

Whenever a net force acts on an object, there is a change in the objects ____