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serious [3.7K]
2 years ago
12

Which Diagram represents summer? ​

Physics
1 answer:
Bingel [31]2 years ago
6 0

Answer:

A

Explanation:

The first diagram best represents summer because the sun faces the earth directly

You might be interested in
A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.
Ivenika [448]

Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

8 0
3 years ago
Una tortuga se desplaza en línea recta 100cm al norte y luego 80cm al este, determine su desplazamiento
valentinak56 [21]

Answer:

is this the full question?

5 0
3 years ago
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0
2 years ago
Try this out. Use a calculator!
Maslowich
Just subsitute and easy
v=55m/s
m=100kg
KE=(0.5)(100kg)(55m/s)^2
KE=(50kg)(3025 m^2/s^2)
KE=151250 J
2nd option
6 0
3 years ago
An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to
Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

\theta_1 is the angle of incidence (in the first medium)

\theta_2 is the angle of refraction (in the second medium)

Let's now consider a situation in which

n_1 > n_2

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

Where \frac{n_1}{n_2} is a number greater than 1. This means that above a certain value of the angle of incidence \theta_1, the term on the right can become greater than 1. So this would mean

sin \theta_2 > 1

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

\theta_c =sin^{-1}(\frac{n_2}{n_1})

8 0
3 years ago
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