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konstantin123 [22]
3 years ago
7

Drop interesting and usefuk facts down here plesse and i'll give the most interesting one brainliest

Physics
1 answer:
Alina [70]3 years ago
4 0
1)The average person spends 6 months of their lifetime waiting on a red light to turn green.
2)16. Men are 6 times more likely to be struck by lightning than women.
3)Coca-Cola would be green if coloring wasn’t added to it
4)Chewing gum while you cut an onion will help keep you from crying.
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As a bullet shot vertically upward rises, the kinetic energy of the bullet
algol [13]

The kenitc energy of the bullet lowers as it keeps going up.

Because gravity is pushing the bullet down as the bullet goes up.

I'm pretty sure that the way to put this answer.

3 0
2 years ago
a fast humvee drove from desert a to desert b. for the 12 hours, it travelled at an speed of 185 km/h. for the next 13 hours, it
ivanzaharov [21]

Answer:

I am not sure of the question?

Explanation:

List the question

8 0
3 years ago
Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

8 0
2 years ago
A cube that has a volume of 1.00 m³ contains N distinguishable particles.
grandymaker [24]

Answer:

1 P = 0.5

2 P = 0.3

3 P = 0.01

Explanation:

The probability formula is

          P =V^N

Where P is the probability  V is the volume while N is the number of distinguishing particles

So for N = 1 and V = 0.500m^3

           P = (0.500m^3)^1\\

               = 0.5

For  N = 1 and V =  0.300m^3

           P = (0.300m^3)^1

               = 0.3

For   N = 1 and V = 0.0100m^3

     P =(0.0100m^3)^1

         = 0.01

         

6 0
2 years ago
A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the
Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

E_i =E_f

so:

Mgh = \frac{1}{2}IW^2 +\frac{1}{2}MV^2

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = \frac{1}{2}MR^2

I = \frac{1}{2}(5kg)(2m)^2

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = \frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2

Replacing the data:

(5kg)(9.8)(0.3m) = \frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2

solving for V:

(5kg)(9.8)(0.3m) = V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))

V = 2 m/s

8 0
3 years ago
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