An automatic response to a stimulus is called a reflex.

Answers

Answer 1
Answer: If you are asking if it is true or false, the correct answer is true. Reflex is an automatic response to a stimulus. Reflex is a stimulus that will allow an individual to respond to an event in which an organism will need to do in order for survival.

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a 1.0 kg ball is thrown into the air with an initial velocity of 30 m/s. How much kinetic energy does the ball have?

Answers

KE=.5mv^2
KE=.5(1)(30^2)
KE=(.5)(900)
KE=450
Assuming no air resistance....

Final answer:

The kinetic energy of a 1.0 kg ball thrown with an initial velocity of 30 m/s is calculated using the formula KE = 1/2 mv^2, resulting in an energy of 450 Joules.

Explanation:

The question you've asked pertains to calculating the kinetic energy of a ball thrown into the air. To find the kinetic energy (KE) of a 1.0 kg ball thrown with an initial velocity of 30 m/s, you can use the formula KE = ½ mv², where m is the mass of the ball and v is the velocity. Plugging in the values, you get KE = ½ × 1.0 kg × (30 m/s)² = 0.5 × 1.0 × 900 = 450 J. Therefore, the kinetic energy of the ball is 450 Joules.

Learn more about kinetic energy here:

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What is the kinetic engery of a baseball five feet above the ground when it has already fallen 32 feet

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The answer is 25 ...

The ratio of output power yo input power, in percent, is called

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The ratio of output power to input power for a machine is called the machine's efficiency. It doesn't have to be expressed in percent.

A flashlight has a resistance of 2.4 what voltage is applied by the batteries if the current in the circuit is 2.5 a?

Answers

Explanation :

The resistance of the flashlight is 2.4 ohms

The current flowing in the circuit is 2.5 A

The Ohm's law gives the relation between the following quantities i.e.

Current

voltage

and resistance

Mathematically, Ohm's law can be written as :

V =I R

V=2.5\ A* 2.4\ \Omega

V=6\ Volts

The voltage applied by the batteries is 6 volts.

Hence, this is the required solution.

Answer:

D - 6.0

Explanation:

i'm not getting more detailed than the other answer so yeah i just took the test

The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). a) True
b) False

Answers

Answer:

b)False

Explanation:

This is the false statement .The two forces in each pair could not have different physical origins.

The two pair of forces have same origin.It can not be possible that one is electric force and other one is gravity force or friction force.

We know that magnitude of electric force is much smaller than the friction force .So they can not be in pair form.

Answer is b)  False

A 0.5 kg ball is dropped from rest at a point 1.2m above the floor. The ball rebounds straight upward to a height of 0.7m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor.

Answers

Answer:

4.281 kgm/s upward

Explanation:

Impulse:This can be defined product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = Change in momentum.

I = mΔv....................... Equation 1.

Where m = mass of the ball, Δv = change in velocity of the ball  

and Δv = v -u

Where u = velocity of the ball before it hit the floor, v = velocity of the ball after if hit the floor

I = m(v-u) -------------- Equation 2

But

the initial kinetic energy of the ball = potential energy at the initial height (1.2 m above)

1/2mu² = mgh₁

Where h₁ = initial height. or height of the ball before collision

making u the subject of the equation,

u = √(2gh₁)........................ Equation 3

Where h₁ = 1.2 m g = 9.81 m/s²

Substitute into equation 3

u = √(2×1.2×9.81)

u =√(23.544)

u = -4.852 m/s.

Note: u is negative because the ball was moving downward at the first instance.

Similarly,

v = √(2gh₂)............................. Equation 3

h₂ = height of the ball after collision

Given: h₂ = 0.7 m, g = 9.81 m/s²

Substitute into equation

v = √(2×9.81×0.7)

v = √13.734

v = 3.71 m/s.

Also given: m = 0.5 kg,

Substituting into equation 2

I = 0.5(3.71-(4.852)

I = 0.5(8.562)

I = 4.281 kgm/s. Upward.

Thus the impulse = 4.281 kgm/s upward