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stepladder [879]
3 years ago
9

The ears are ____ and ____ to the shoulders and ____ to the nose.

Physics
2 answers:
Levart [38]3 years ago
3 0
The ears are superior and posterior to the shoulders and lateral to the nose.
LiRa [457]3 years ago
3 0

Which describes a relationship based on the general theory of relativity?

Acceleration in space and magnetism on Earth have opposite effects on objects.Acceleration in space and magnetism on Earth have the same effects on objects.Acceleration in space and gravity on Earth have opposite effects on objects.<span>Acceleration in space and gravity on Earth have the same effects on objects. </span>
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Heating food under a heat lamp is an example of heat transfer by
Tcecarenko [31]
<span>Heating food under a heat lamp is an example of heat transfer by <span>Radiation</span></span>
5 0
3 years ago
A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m
Feliz [49]

Explanation:

Let the speeds of father and son are v_f\ and\ v_s. The kinetic energies of father and son are K_f\ and\ K_s. The mass of father and son are  m_f\ and\ m_s

(a) According to given conditions, K_f=\dfrac{1}{3}K_s

And m_s=\dfrac{1}{4}m_f

Kinetic energy of father is given by :

K_f=\dfrac{1}{2}m_fv_f^2.............(1)

Kinetic energy of son is given by :

K_s=\dfrac{1}{2}m_sv_s^2...........(2)

From equation (1), (2) we get :

\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}

v_s^2=4(v_f+1.5)^2

Using equation (3) in above equation, we get :

v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s

(b) Put the value of v_f in equation (3) as :

v_s=7.09\ m/s

Hence, this is the required solution.

8 0
3 years ago
What are the basic si units for the speed of light?.
DanielleElmas [232]
C=meters/second or C=m/s
7 0
2 years ago
A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
Mandy is testing an unknown solution to determine whether it is an acid or a base. She places a piece of red litmus paper into t
Marina CMI [18]
The solution is a base
5 0
3 years ago
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