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Akimi4 [234]
3 years ago
10

(answer only if you know the answer or I'll report) Help me please solve it w steps​

Physics
1 answer:
Murrr4er [49]3 years ago
5 0

X component of force    Fx=Fcos45=50×cos×45

Y component of force    Fy=Fsin45=50×sin×45

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In 1996, astronomers discovered an icy object beyond pluto that was given the designation 1996 tl 66. it has a semimajor axis of
antoniya [11.8K]

Answer : 2446 years.

Explanation :

Length of semi major axis is, a=84\ au= 1.496\times 10^{11}\ m

According to Kepler's third law, square of time period of an orbit is directly proportional to the cube of the semi major axis.

i.e T^2=\dfrac{4\pi^2}{GM}a^3

where G is gravitational constant

M is  mass of sun, M=1.98\times 10^{30}\ Kg

So, T^2=\dfrac{4\times (3.14)^2}{6.6\times 10^{-11}Nm^2/Kg\times 1.98\times 10^{30}\Kg}

T^2=3\times 10^{-19}\times(84\times 1.496\times 10^{11})^3

T^2=3\times 10^{-19}\times 1984415.6\times 10^{33}

T^2=59532469.8\times 10^{14}\ s

T=7715.7\times 10^7\ s

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7 0
4 years ago
Select the correct answer.
Nina [5.8K]

Answer:

Ос.

The spacing between particles increases.

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3 0
3 years ago
A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.
Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0
3 years ago
You lift a 45-newton bag of mulch 1.2 meters and carry it a distance of 10 meters to the garden. How much 
zysi [14]
To lift the bag straight up takes (F · D) = (45 · 1.2) = 54 joules of energy (work).

Moving the bag horizontally 'across' gravity requires no work.
It doesn't matter how far.
4 0
4 years ago
True or false
madam [21]
That's true, you can't get any more precise than what's measured to the least precision.
8 0
3 years ago
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