Which statement describes the vector plotted below?

marissas car accelerates uniformly at a rate of -2.60m/s^2. how long does it take for marissas car to accelerate from a velocity

zlopas [31]

Answer:

1.02s

Explanation:

In this situation the following equation will be useful:

Where:

is Marissa's car final velocity

is Marissa's car initial velocity

is Marissa's car constant acceleration (assuming this is the acceleration, since 1269 m/s^{2} does not make sense)

is the time it takes to accelerato from to

7 0

2 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizon

kirza4 [7]

Answer:

A) 3.79 m/s B) 1.33 m

Explanation:

A)

Horizontal movement:
Once in the air, no forces act on the froghopper, so it keeps moving with the same initial horizontal speed.
This horizontal component, is the projection of the velocity vector on the horizontal direction (x-axis):

$v_{ox} = v_{o} *cos (58.0 deg)$

The horizontal displacement can be simply calculated as follows:

$x = v_{ox} *t$

Vertical movement:
As the vertical and horizontal are independent each other (due to they are perpendicular, so there is no projection of one movement on the other), in the vertical direction, all happens as if would be a body thrown upward with a given initial vertical velocity.
This velocity can be found as the projection of the velocity vector on the vertical direction (y-axis):

$v_{oy} = v_{o} *sin (58.0 deg) (1)$

Once in the air, the gravity will cause that the froghopper be slow down, till it reaches to the maximum height, where it will come momentarily to an stop.
In that moment, we can apply the following kinematic equation:

$v_{fy} ^{2} -v_{oy} ^{2} = 2*g*h_{max}$

where vfy = 0, g = -9.8m/s2, hmax = 52.7 cm= 0.527 m
Replacing by the givens, we can solve for voy:

$v_{oy} =\sqrt{2*g*h_{max}} = \sqrt{2*9.8m/s2*0.527m} =3.21 m/s$

From the equation (1), we can solve for the magnitude of the initial velocity, v₀:

$v_{o} = \frac{v_{oy}}{sin 58.0} =\frac{3.21m/s}{0.848} = 3.79 m/s$

B)

With the value of the magnitude of the initial velocity, we can find the horizontal component vox, as follows:

$v_{ox} = v_{o} *cos (58.0 deg) =\\ \\ 3.79 m/s * cos (58.0deg) = 2.01 m/s$

In order to know the horizontal distance travelled, we need to find the time that the insect was in the air.
We can use the equation for the vertical displacement, replacing this value by 0, as follows:

$y = 0 = v_{oy} *t -\frac{1}{2} * g *t^{2}$

Replacing by the givens, and rearranging terms, we can solve for t:

$t_{air} =\frac{2*v_{oy} }{g} = \frac{2*3.21 m/s}{9.8 m/s} = 0.66 s$

Finally, we find the horizontal displacement, as follows:

$x_{max} = v_{ox} *t_{air} = 2.01 m/s * 0.66 s \\ \\ x_{max} = 1.33 m$

The horizontal distance covered by the froghopper was 1.33 m.

4 0

3 years ago

Which is NOT a subatomic particle

Novay_Z [31]

Answer:

both proton and neutron are subatomic particle

Explanation:

hope this helps u stay safe

6 0

3 years ago

Which data set has the largest standard deviation

AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

$\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}$