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Keith_Richards [23]
3 years ago
11

Is it B? Not too sure about this one.

Physics
2 answers:
klasskru [66]3 years ago
7 0
no the correct answer is C
olchik [2.2K]3 years ago
3 0
The correct answer to this problem is C
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What branch of science that deals with landforms,features,inhabitants and phenomena on earth
iren2701 [21]

Answer:

Earth and space

Explanation:

4 0
3 years ago
A basketball rolls without slipping (starting from rest) down a ramp. If the ramp is sloped by an angle of 4 degrees above the h
slavikrds [6]

Answer:

11.7 m/s

Explanation:

To find its speed, we first find the acceleration of the center of mass of a rolling object is given by

a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²

a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)

  = 9.8 m/s²sin4(1 + 2/3)

  = 9.8 m/s²sin4 × (5/3)

  = 1.14 m/s²

To find its speed v after rolling for 60 m, we use

v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m

v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s

4 0
3 years ago
What do you think we call this graphical representation based on your prior experience with electric fields and electric field l
slavikrds [6]

Answer:

Explanation:

The strengthcompassion field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. The direction of the electric field is tangent to the field line at any point in space. Field lines can never cross. These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are directed away from positively charged source charges and toward negatively charged source charges.

Rules for drawing electric field lines

1. Electric field lines are always drawn from High potential to

low potential.

2. Two electric field lines can never intersect each other.

3. The net electric field inside a Conductor is Zero.

4. Electric field line from a positive charge is drawn radially outwards and from a negative charge radially inwards.

5. The density of electric field lines tells the strength of the electric field at that region.

6. Electric field lines terminate Perpendicularly to the surface of a conductor.

A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.

So, electric fields are vector quantity due to the fact any student can tell you that a compass is used to determine which direction is north.

Since the compass always point northward, then it has a direction and magnitude and so it is a vector quantity

6 0
3 years ago
A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15
cupoosta [38]

Answer:

\boxed {\boxed {\sf 4.5 \ m/s \ in \ the  \ positive \ direction}}

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

v_f= v_i + at

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

  • v_i= 2.7 m/s
  • a= 0.15 m/s²
  • t= 12 s

Substitute the values into the formula.

v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)

Multiply the numbers in parentheses.

v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)

v_f = 2.7 \ m/s + (0.15 \ m/s *12)

\v_f=2.7 \ m/s + (1.8 \ m/s)v_f=2.7 \ m/s + (1.8 \ m/s)

Add.

v_f=4.5 \ m/s

The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>

5 0
2 years ago
Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a
cupoosta [38]

Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point =  mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

7 0
3 years ago
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