Speed is a component of skill related fitness what does speed enable you to do.

Explain the why a constant speed has a slope of 0 on a graph of speed v.time.

Studentka2010 [4]

Because if I do 0*time you get the answer as 0

6 0

3 years ago

Which of the following would illustrate a quadratic relation between the dependent and independent variables when graphed?

Kitty [74]

Answer: option A. a graph of the area of a circle vs. its radius r (A = πr²).

Explanation:

A quadratic relation between the dependent and independent variables shows the independent variable raised to the power of 2.

This is it is a polynomial with general form ax² + bx + c, whewre a, b, and c, named coeficients, are constants.

The function is y = ax² + bx + c, where x is the independent variable and y is the dependent variable.

As stated in the question, the area of a circle is given by A = πr².

In this case, A is the dependent variable and r is the independent variable.

π is assumed as the coefficient of the quadratic term, and the other coefficients are assumed 0, since there are no either terms on r or constants.

The equation a = 1/b is an inverse relation, not a quadratic relation.

The relation of distance vs. time for a car moving at constant speed is a linear relation of the kind v = u + st.

The mass of water vs. the volume of water in a drinking glass is a direct relation, mass = density × volume

Therefore, the only quadratic relation is shown by a graph of the area of a circle vs. its radius r.

3 0

2 years ago

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).

5 0

3 years ago

Which statement best summarizes the central idea of “Applications of Newton’s Laws”?

Shkiper50 [21]

Newton's three forces, normal, tension and friction, are present in a surprising number of physical situations

Newton's Laws, that describe the relationship between an obejct and the forces acting upon it, apply in almost every physical situation, from quantum mechanics to electricity.

The correct answer is:

Newton’s laws can explain the forces that occur between objects every day

3 0

3 years ago

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