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3 years ago

In Linnean taxonomy, the most unique level is _____. species genus kingdom breed

Physics

2 answers:

Nookie1986 [14]3 years ago

6 0

The answer is species, I hope this helps!

dezoksy [38]3 years ago

5 0

The correct answer is A I'm doing the quiz too

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In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

2 years ago

Question 9 of 10

mixer [17]

Answer:

I believe the correct answer is B..<u>" 1 "</u>

Explanation:

hope it helps! ;)

8 0

2 years ago

What’s a substance made up of two or more elements

Alex

Answer:

That would be a molecule. : )

Explanation:

The definition of a molecule is a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.

7 0

2 years ago

A 62 kg boy and a 37 kg girl use an elastic rope while engaged in a tug-of-war on a friction-less icy surface. If the accelerati

Paraphin [41]

Answer:

The magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

Explanation:

It is given that,

Mass of the boy, $m_1=62\ kg$

Mass of the girl, $m_2=37\ kg$

The acceleration of the girl toward the boy is, $a_2=2.2\ m/s^2$

To find,

The acceleration of the boy toward the girl.

Solution,

Let $a_1$ is the magnitude of the acceleration of the boy toward the girl. We know that force acting on one object to other are equal in magnitude but opposite in direction. So,

$F_1=-F_2$

$m_1a_1=-m_2a_2$

$a_1=-\dfrac{m_2a_2}{m_1}$

$a_1=-\dfrac{37\times 2.2}{62}$

$a_1=-1.31\ m/s^2$

$|a_1|=1.31\ m/s^2$

So, the magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

7 0

2 years ago

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pochemuha

The motion of the buoy is a composition of two independent motions:
- a uniform motion on the horizontal axis, with constant speed vx=50 m/s
- an uniformly accelerated motion on the vertical axis, with constant acceleration $g=9.81 m/s^2$

Since we want to find the vertical displacement, we are only interested in the vertical motion.
The law of motion on the vertical direction is given by:
$y(t)=h+v_{0y} t+ \frac{1}{2}gt^2$
where
h is the initial height of the buoy
$v_{0y}$ is the initial vertical velocity of the buoy, which is zero
t is the time

We know that the buoy lands after t=21 seconds, this means that the vertical position at t=21 s is y(21 s)=0. If we substitute these data into the equation, we can find the value of h, the initial height of the buoy:
$0=h+ \frac{1}{2}gt^2$
$h= -\frac{1}{2}gt^2= -\frac{1}{2}(9.81 m/s^2)(21 s)^2=-2163 m$
And this corresponds to the vertical displacement of the buoy.

7 0

3 years ago