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3 years ago

Use complete sentences to describe an extra-solar (exo) planet

Physics

2 answers:

Jobisdone [24]3 years ago

5 0

An exo planet is a planet that orbits a stat outside of the solar system.

anyanavicka [17]3 years ago

4 0

The correct answer of this question : The planet outside of our solar system.

EXPLANATION:

The extra-solar planet is the planet which is present beyond our solar system. It is also known as exoplanet.

Just like our solar system planets like the earth,Mars, Jupiter etc, the exoplanet is also a part of another solar system. The planets revolves around their sun i.e other star in different orbits. This star is the centre of their solar system.

Hence, the exact explanation for the exoplanet is the planet situated beyond our solar system, and revolves around any other centrally situated star.

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Two forces, F1 and F2 (with F1 > F2), act on an object of mass m. The two forces act along the same line but in opposite dire

natali 33 [55]

Here we can use Newton's II law to find the acceleration of the car

$F_{net} = ma$

here we know that

$F_{net} = F_1 - F_2$

as we know that two forces are equal in magnitude but opposite in direction so in order to find net force we need to subtract it

now from above equation

$F_1 - F_2 = ma$

$a = \frac{(F_1 - F_2)}{m}$

so here option A is correct

5 0

3 years ago

A pair of scissors is a complex machine. what simple machines make up a pair of scissors?

vova2212 [387]

2 blades and a securing bolt.

4 0

3 years ago

Read 2 more answers

A crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and

lina2011 [118]

Answer:0.3

Explanation:

Given

velocity of car=15 m/s

truck brought to halt in a distance of 38 m

We know

$v^2-u^2=2as$

Final velocity (v)=0

$0-(15)^2=2(a)(38)$

$a=\frac{-225}{76}$

$a=-2.96 m/s^2$ (deceleration)

Therefore minimum coefficient of friction \mu will be

$\mu \times g=a$

$\mu =\frac{a}{g}$

$\mu =\frac{2.96}{9.8}=0.302$

7 0

3 years ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

lbvjy [14]

Answer:

Electric flux;

Φ = 30.095 × 10⁴ N.m²/C

Explanation:

We are given;

Charge on plate; q = 17 µC = 17 × 10^(-6) C

Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²

Angle between the normal of the area and electric field; θ = 4°

Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m

Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²

The charge density on the plate is given by the formula;

σ = q/A_p

Thus;

σ = (17 × 10^(-6))/(180 × 10^(-4))

σ = 0.944 × 10^(-3) C/m²

Also, the electric field is given by the formula;

E = σ/ε_o

E = (0.944 × 10^(-3))/(8.85 × 10^(-12))

E = 1.067 × 10^(8) N/C

Now, the formula for electric flux for uniform electric field is given as;

Φ = EAcos θ

Where A = πr² = π × 0.03² = 9π × 10^(-4) m²

Thus;

Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4

Φ = 30.095 × 10⁴ N.m²/C

3 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

lina2011 [118]

Answer:

-6.0 m/s, 10.4 m/s

Explanation:

To find the x- and y- components, we have to apply the formulas:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v = 12.0 m/s is the magnitude of the vector

$\theta$ is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

$\theta=180^{\circ} - 60^{\circ} = 120^{\circ}$

So, the two components are:

$v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s$

$v_y = (12.0 m/s) sin 120^{\circ}=10.4 m/s$

5 0

3 years ago