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3 years ago

Nodes give a vertebrate’s back flexibility. A. True B. False

Physics

2 answers:

Elis [28]3 years ago

8 0

B- False it
invertebrates have flexibility ( octopuses and squids are good example of invertebrates)

Lapatulllka [165]3 years ago

5 0

No B. its false<span> they do not. Lyphnodes are a permant part of the body and you have them in your Neck</span>

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An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q

Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency $f = ( \dfrac{1}{2 \pi}) \omega$

where;

$\omega = \sqrt{\dfrac{k}{m} }$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )$

However;

$k = \dfrac{F}{x}$ and;

mass $m = m_{car } + m_{person}$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )$

where;

$F = m_{person}g$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )$

replacing the values;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )$

$\mathbf{f = 0.442 \ Hz}$

8 0

3 years ago

What is the mass of an object that requires 100N of force in order to accelerate it at 10m/s2 ?

Debora [2.8K]

Answer:

We know that f=ma

Rearranging

M=f÷a so m=10kg

Explanation:

7 0

3 years ago

Read 2 more answers

In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a

katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

brainly.com/question/11422992

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3 0

2 years ago

You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from

maxonik [38]

Answer:

A. There is a localization of positive charge near the door handle.

Explanation:

When on a cold morning a person wearing cotton/ polyester cloth walking on the carpet moves toward his car then due to friction between the feet and the carpet there are transfer of electrons from the carpet to our feet, and since our body is a good conductor of electricity the charges spread throughout on the surface of or body.

When the person brings his hands close to the neutral conducting door of the car it gets induced with equal intensity of opposite charge to our hands thus having a concentration of positive charges near to the hand on the car's door is developed as a result of polarization within the conductor.

3 0

3 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$