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Scrat [10]
3 years ago
8

Where are fossil fuels found?

Physics
2 answers:
astraxan [27]3 years ago
6 0

Answer:

B. in the Earth

Explanation:

Fossil fuels are found in the Earth's crust and contain carbon and hydrogen, which can be burned for energy.

Brainlist pls!

Annette [7]3 years ago
5 0

Answer:

in earth

eghdksksbbxmdowosbb

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A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
PtichkaEL [24]

Explanation:

(a) Find the magnitude of the angular acceleration of the wheel.

  • angular acceleration = angular speed /time
  • angular acceleration = 12.9/2.98 = 4.329rad/s²

(b) Find the angle in radians through which it rotates in this time interval.

  • angular speed = 2x3.14xf
  • 12.9rad = 2 x3.14

  • rad = 6.28/12.9
  • rad = 0.487

Now we convert rad to angle

  • 1 rad = 57.296°
  • 0.487 = unknown angle
  • unknown angle =57.296 x 0.487 = 27.9°

The angle in radians = 27.9°

8 0
3 years ago
Why does an astronaut in a spacecraft orbiting Earthexperience<br> a feeling of weightlessness?
slava [35]

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.                      

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

4 0
3 years ago
Which galaxy is the most stretched out?<br>​
RideAnS [48]

Answer:

this is the anwser

Explanation:

The oddball spiral galaxy, called Messier 66, is one-thirdof the Leo Triplet, a group of three interacting galaxies about 35 millionlight-years from Earth (a light-year is the distance light can cover in ayear).

4 0
3 years ago
Read 2 more answers
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)
Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is A_{x} = 0

the y-component of the force at A is A_{y}  = 400 N

the couple acting at A is; M_{A} = 146 N-m

b)

the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑M_{L} = 0

Vector force acting at A, F_{A} = A_{x}i + A_{y}j

Now, From the image, we have;

a)

∑f = 0

F_{A} - 600j + 200j = 0i + 0j

A_{x}i + A_{y}j - 600j + 200j = 0i + 0j

A_{x}i + (A_{y} - 400)j = 0i + 0j

now by equating i- coefficients'

A_{x} = 0

so, the x-component of the force at A is A_{x} = 0

also by equating j-coefficient

A_{y} - 400 = 0

A_{y}  = 400 N

hence, the y-component of the force at A is A_{y}  = 400 N

we also have;

∑M_{L} = 0

M_{A}  - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

M_{A} - 30 N-m - 228 N-m + 112 Nm = 0

M_{A} - 146 N-m = 0

M_{A} = 146 N-m

Therefore, the couple acting at A is; M_{A} = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑M_{R} = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(A_{y} ) + M_{A}

∑M_{R} = (108  N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑M_{R} = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑M_{R}  = 0

Therefore, the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

7 0
2 years ago
If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w
Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
5 0
3 years ago
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