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aleksandrvk [35]
3 years ago
15

How many regions does the spine have and what are them ?s

Physics
1 answer:
Viktor [21]3 years ago
4 0

the spine is divided into four main regions: cervical, thoracic, lumbar and sacral.

Have an amazing day! <3

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What is the total number of electrons in the outermost shell of a phosphorus atom in ground state?
Leya [2.2K]
The answer is 3 I hope I helped
3 0
4 years ago
What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25
Rom4ik [11]

Acceleration of the ball is 50 m/s^2

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2

Learn more about Newton's second law:

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5 0
3 years ago
Is molten lava heterogeneous
zhenek [66]
Yes, considering lava accommodates rocks that have collapsed of cave walls it is heterogeneous. 
4 0
3 years ago
A small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then dro
Lunna [17]

• The speed change between T and S is greater between S and R.

• The speed of m at X is  greater that at Q.

• The size of the total force on m at P is less at U.

• The mechanical energy of m at P is equal to that at V.

• The size of the total force on m at S is greater at P.

• The velocity of m at X is equal to that at V.

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.

The picture shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V.

Between T to S and S to R, the mass is under constant acceleration. Time taken to move from T to S is greater than S to R. Thus, the speed change between T and S is greater than between S and R.

At Q, there is only a horizontal velocity component, but at X. the speed will be greater and has both vertical and horizontal component. Thus, the speed of m at X is greater than that at Q.

Force is given as the rate of change of momentum with time. At U, change in momentum is large compared to P. Thus, the size of the total force on m at P is less at U.

There is no friction acting on the system. So the energy remains conserved. Mechanical energy at P = V.

The force on mass m at S is only the gravity force. The remaining forces are cancelled by the normal force. Thus, size of the total force on m at S is greater at P.

The energy is conserved at each point of motion of mass. If X and V are at same height, they have same potential energy and so their kinetic energy. Thus, velocity of m at X is equal to that at V.

Learn more about mechanical energy.

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5 0
2 years ago
A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

5 0
3 years ago
Read 2 more answers
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