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KatRina [158]
3 years ago
9

the scores of players on a golf team are shown in the table. the teams combined score was 0 what was travis's score?

Physics
2 answers:
barxatty [35]3 years ago
7 0

Answer:

-5

Explanation:

Alona [7]3 years ago
3 0

Answer:

what table?

Explanation:

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I need an answer ASAP .....
Harrizon [31]

Answer:

Hope this helped :

Explanation:

When light passes from a medium with one index of refraction (m1) to another medium with a lower index of refraction (m2), it bends or refracts away from an imaginary line perpendicular to the surface (normal line). As the angle of the beam through m1 becomes greater with respect to the normal line, the refracted light through m2 bends further away from the line.

At one particular angle (critical angle), the refracted light will not go into m2, but instead will travel along the surface between the two media (sine [critical angle] = n2/n1 where n1 and n2 are the indices of refraction [n1 is greater than n2]). If the beam through m1 is greater than the critical angle, then the refracted beam will be reflected entirely back into m1 (total internal reflection), even though m2 may be transparent!

4 0
3 years ago
Read 2 more answers
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0
3 years ago
Read 2 more answers
What is the magnitude of the magnetic dipole moment of the bar magnet
Annette [7]

The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²

<h3> Magnetic dipole moment of the bar magnet</h3>

The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

B = \frac{2\mu_0m}{4\pi r^3} \\\\m = \frac{4\pi r^3 B}{2\mu_0}

where;

  • B is magnetic field
  • m is dipole moment
  • μ is permeability of free space

m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)

m = 1.2 Am²

The complete question is below:

What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.

Learn more about dipole moment here: brainly.com/question/27590192

#SPJ11

6 0
2 years ago
Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a
Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

          F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

        P = (m1 + m2) a

        a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block.  In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

       N = m2 a

       fr -W2 = 0    

       fr = W2

       

The definition of friction force is

       fr = μ N

       

Let's replace and calculate

       μ (m2 a) = m2 g

       μ (P / (m1 + m2)) = g

       P = g /μ  (m1 + m2)

Let's calculate the value of this force

       P = 9.8 / 0.710 (28.9 +4.4)

       P = 13.80 (33.3)

       P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0
3 years ago
Science: Please help me with this due in 5 minutes
babunello [35]

Answer: Use less water

only turn on lights when needed

electrical cars

less gas usage

Explanation:

6 0
3 years ago
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