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meriva
3 years ago
12

What is same about all myriapods

Physics
1 answer:
seraphim [82]3 years ago
4 0
They all have segmented limbs, a hard exoskeleton, a pair of antennae and a segmented body.
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An astronaut has a momentum of 280 kg and travels 10 m/s. what is the mass of the astronaut?
Kamila [148]

Answer:

The answer is

<h2>28 kg</h2>

Explanation:

The mass of an object given it's momentum and velocity / speed can be found by using the formula

m =  \frac{p}{v}  \\

where

m is the mass

p is the momentum

v is the speed or velocity

From the question

p = 280 kg/ms

v = 10 m/s

The mass of the object is

m =  \frac{280}{10}  = 28 \\

We have the final answer as

<h3>28 kg</h3>

Hope this helps you

3 0
3 years ago
Which of Newton's Three Laws Applies? Law 1, 2, or 3?
frozen [14]

Answer:

1. Newton's first law

2.Newton's second law

3.Newton's third law

Explanation:

1. Newton's first law stated, In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force... this is base of the concept of inertia.

2. Newton's second law stated, In an inertial frame of reference, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma, or in easier words, F is directly proportional to a.

3. Newton's third law stated, When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body., In this case, the Normal Are opposite with gravititional force.

7 0
3 years ago
Gabriella got a burn on her arm from a chemical spill and rinsed the area thoroughly with water. Which piece of safety equipment
ioda
I believe it she should use the first aid kit next
8 0
3 years ago
Read 2 more answers
An object has a mass of 20 g and a volume of 5 cm3. what is the object's density? 4 g/cm3 100 g/cm3 0.25 g/cm3
Sati [7]
The answer is 110 g/cm3
8 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
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