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4 years ago

Which two pieces of data indicate that Uranus resides in the outer region of the solar system

Physics

1 answer:

LuckyWell [14K]4 years ago

4 0

Answer:

Our solar system has total eight planets out of which four are inner planets and four are outer planets. The four outer planets are Jupiter, Saturn, Uranus and Neptune. The common characteristics of outer planets is that they are gaseous planets. They are larger on size than the inner rocky planets and are faraway from Sun. They have larger period of revolution around the Sun.

Uranus is a gaseous planet and lies far from Sun and hence has large period of revolution. It takes 84 Earth years to revolve around Sun. This data indicates that Uranus resides in the outer region of the Solar System.

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A bowling ball has a mass of 6 kilograms. A person lifts the bowling ball 2 meters above the ground, using a force of about 60

Luba_88 [7]

The amount of work done by the person is equal to 120 Joules.

<u>Given the following data:</u>

Force = 34 Newton.

Displacement = 15 meters.

Mass = 6 kilograms.

To calculate the amount of work done by the person at a total height of 2 meters:

<h3>How to calculate work done.</h3>

In Science, work done is calculated by multiplying force and the displacement experienced by an object.

Mathematically, work done is given by the formula:

$Work\;done = force \times displacement$

Substituting the given parameters into the formula, we have;

$Work\;done = 60 \times 2$

Work done = 120 Joules.

Read more on work done here: brainly.com/question/22599382

4 0

2 years ago

Which of the following best describes reflection?

Margarita [4]

B. Reflection is when waves bounce off a surface Ik this because I did it in 7th grade was my favorite unit to learn in school

4 0

3 years ago

Read 2 more answers

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

sergij07 [2.7K]

Answer:

A) 50.0 g Al

Explanation:

We can calculate the temperature change of each substance by using the equation:

$\Delta T=\frac{Q}{mC_s}$

where

Q = 200.0 J is the heat provided to the substance

m is the mass of the substance

$C_s$ is the specific heat of the substance

Let's apply the formula for each substance:

A) m = 50.0 g, Cs = 0.903 J/g°C

$\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C$

B) m = 50.0 g, Cs = 0.385 J/g°C

$\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C$

C) m = 25.0 g, Cs = 0.79 J/g°C

$\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C$

D) m = 25.0 g, Cs = 0.128 J/g°C

$\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C$

E) m = 25.0 g, Cs = 0.235 J/g°C

$\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C$

As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.

7 0

3 years ago

Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6

inna [77]

Answer:

Answer:

Q_1 = 7Q

Q_2 = 10Q

=10

Q_3 = 13.5Q

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

(9+1)

Q_1 = \frac{1}{4}*10Q

∗10

Q_1 = 2.5th\ itemQ

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

(7+7)

Q_1 = \frac{1}{2}*14Q

∗14

Q_1 = 7Q

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

(9+1)

Q_2 = \frac{1}{2}*10Q

∗10

Q_2 = 5th\ itemQ

=5th item

Q_2 = 10Q

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

(9+1)

Q_3 = \frac{3}{4}*10Q

∗10

Q_3 = 7.5th\ itemQ

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

(12+15)

Q_3 = \frac{1}{2}*27Q

∗27

Q_3 = 13.5Q

=13.5

4 0

3 years ago

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses

Tom [10]

Answer:

a) U = 0.375 mJ

b) p_rad = 4.08 mPa

c) λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d) E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

Explanation:

Given:

λ_air : wavelength = 810 * 10^(-9) m

P: Power delivered = 0.25 W

d : Diameter of circular spot = 0.00051 m

c : speed of light vacuum = 3 * 10^8 m/s

n_air : refraction Index of light in air = 1

n_med : refraction Index of light in medium = 1.34

ε_o : permittivity of free space = 8.85 * 10^-12 C / Vm

part a

The Energy delivered to retina per pulse given that laser pulses are 1.50 ms long:

U = P*t

U = (0.25 ) * (0.0015 )

U = 0.375 mJ

Answer : U = 0.375 mJ

part b

What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed?

p_rad = I / c

Where I : Intensity = P / A

p_rad = P / A*c

Where A : Area of circular spot = pi*d^2 / 4

p_rad = 4P / pi*d^2*c

p_rad = 4(0.25) / pi*0.00051^2*(3.0 * 10^8)

p_rad = 0.00408 Pa

Answer : p_rad = 4.08 mPa

part c

What are the wavelength and frequency of the laser light inside the vitreous humor of the eye?

λ_med = n_air*λ_air / n_med

λ_med = (1) * (810 nm) / 1.34

λ_med = 604 nm

f_med = f_air

f_med = c / λ_air

f_med = (3*10^8) / (810 * 10^-9)

f_med = 3.7 * 10^14 Hz

Answer : λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

What is the electric and magnetic field amplitude in the laser beam?

I = P / A

I = 0.5*ε_o*c*E_o ^2

I = 4P / pi*d^2

Hence, E_o = ( 8 P / ε_o*c*pi*d^2 ) ^ 0.5

E_o = ( 8 * 0.25 / (8.85*10^-12) * (3*10^8) * π * (0.00051)^2) ^ 0.5

E_o = 3.04 * 10^4 V / m

For maximum magnetic field strength:

B_o = E_o / c

B_o = 3.04 * 10^4 / (3*10^8)

B _o = 1.013 *10^-4 Nm/Amp

Answer: E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

5 0

3 years ago