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luda_lava [24]
2 years ago
9

Which statement describes the density of the inner planets?

Physics
1 answer:
Eduardwww [97]2 years ago
5 0

Answer:

The inner planets are less dense than the outer planets

Please brainliest my answer iam begging you for god sake

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The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu
abruzzese [7]

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6.8 \times  {5}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3.4 \times 25

\sf \implies KE =3.4 \times 25

\sf \implies KE = 85 \: J

8 0
3 years ago
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I’ll give brainliest if it’s correct ;-;z
BlackZzzverrR [31]

Explanation:

what is the question? could you pls provide it

6 0
2 years ago
Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh
coldgirl [10]

Answer:

You cant draw on brainly

Explanation:

6 0
2 years ago
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URGENT!!! DUE AT 11:59, PLEASE DON’T POST A LINK FOR THIS ANSWER AND BE STRAIGHT FORWARD! Bronco the skydiver, whose mass is 100
kirill [66]

a = 7.8 m/s^2

Explanation:

Let Fnet = net force = ma

m = mass of the skydiver

a = acceleration caused by Fnet

W = weight = mg

f(air) = frictional force due to air resistance

Fnet = W - f(air)

= (100 kg)(9.8 m/s^2) - (200 N)

= 780 N

Therefore, the acceleration of the skydiver due to Fnet is

a = Fnet/m

= (780 N)/(100 kg)

= 7.8 m/s^2

4 0
3 years ago
A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.
Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0
3 years ago
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