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katen-ka-za [31]
3 years ago
11

What is this answer

Physics
2 answers:
Katyanochek1 [597]3 years ago
7 0
Answer: an ectomorph is a body type that struggles to gain weight and muscle
gulaghasi [49]3 years ago
3 0
The answer is To gain weight and Muscle.
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How much of the universe do scientists speculate is composed of dark energy
algol [13]

Answer: Approximately 65% from what i have learnt.

5 0
3 years ago
Convert the speed of light 3.0x10^8 m/s to km/day
Aleksandr [31]

Answer: 2.592 \times 10^{8}km/day

1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day



7 0
3 years ago
Read 2 more answers
You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient
iren [92.7K]

Answer:

angle minimum   θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

    N - W = 0

    N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

     fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

     

the friction force formula is

     fr = μ N

     fr = μ W

we substitute

      μ m g l sin θ - m g l cos θ + mg l /2   cos θ = 0

      μ sin θ - cos θ + ½ cos θ= 0

         

       μ sin θ - ½ cos θ = 0

       sin θ / cos θ = 1/2 μ

       tan θ = 1/2 μ

       θ = tan⁻¹ (1 / 2μ)

       θ = tan⁻¹ (1 (2 0.57))

      θ = 41.3º

7 0
3 years ago
The outer planets are similar to the planet Earth; they just happen to be farther away from the Sun.
Harrizon [31]

Answer:

False

Explanation:

The inner planets are called terrestrial planets due to the surfaces are solid (similar to Earth)-made up of heavy metals, either have no moons or few moons.

The outer planets are called Jovian planets or gas giants because they are encased in gas.  They all have rings with plenty of moons.

5 0
3 years ago
Read 2 more answers
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
4 years ago
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