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svp [43]
2 years ago
15

20. Which of the following situations is reaction time exhibited?

Physics
1 answer:
sashaice [31]2 years ago
4 0

Jenny and Tina play jumping rope,Ben and Ted plays soccer and football and Bovet and Cris perform aerobic exercises​ are the following situations is reaction time exhibited. Option A,B and D are correct.

<h3>What is reaction time?</h3>

The reaction time of an organism is a measurement of how quickly it responds to a stimulus.

The time interval between the presentation of the stimulus and the manifestation of a suitable voluntary response in the subject is defined as RT.

In the following situations is reaction time exhibited is;

A. Jenny and Tina play jumping rope.

B. Ben and Ted play soccer and football .

D. Bovet and Cris perform aerobic exercises​.

Hence,option A,B and D are correct.

To learn more about reaction time refer:

brainly.com/question/13693578

#SPJ1

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A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length
Ad libitum [116K]

Answer:

a)  I = 0.0198 kg m² ,  b)    I = 21.85 kg m²

Explanation:

For this exercise we will use the definition of moment of inertia

        I = ∫ r² dm

For body with high symmetry they are tabulated

sphere  I = 2/5 m r²

bar with respect to  center of mass I = 1/12 m L²

let's calculate the mass of each body

bar

        ρ = m / V

        m = ρ V

        m = ρ l w h

where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm

        m = 32820 1 0.008 0.04

        m = 10.5 kg

Sphere

       M = ρ V

       V = 4/3 pi r³

       M = rgo 4/3 π r³

give us the density 37800 kg / m³ and the radius of 5 cm

       M = 37800 4/3 π 0.05³

       M = 19.8 kg

a) asks us for the moment of inertia of the sphere with respect to its center of mass

        I = 2/5 M r²

        I = 2/5 19.8 0.05²

        I = 0.0198 kg m²

b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes

        I = I_cm + M d2

where d is the distance from the body to the point of interest

        I_cm = 0.0198 kg m²

the distance to the pivot point is

        l = length of the bar + radius of the sphere

        l = 1 + 0.05 = 1.005 m

        I = 0.0198 + 19.8 1.05²

        I = 21.85 kg m²

8 0
3 years ago
A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.
natima [27]

Answer:

(a)  24.025 m/s. downward.

(b)  31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

Constant : g = 9.81 m/s²

Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

s₁ = -1.25+30.656

s₁ = 29.41 m

also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

s₂ = 0.5(2.5)

s₂ = 1.25 m.

Note: Distance between the pelican and fish = s₁ + s₂

Distance between the pelican and fish  = 29.41+1.25

Distance between the pelican and fish  = 30.66

Distance between the pelican and fish ≈ 31 m

3 0
3 years ago
Graphing data is most used to help you to observe_____ *
kotykmax [81]

Answer: Graphing data is used to display data because it is easier to see trends in the data when it is displayed visually compared to when it is displayed numerically in a table.

5 0
3 years ago
Two particles with charges of 5.00 μ C and -3.00 μC are placed 0.250 m apart. Where can a third charge be placed so that the net
MAXImum [283]

Answer:

0.86 m

Explanation:

q₁ = magnitude of positive charge = 5 x 10⁻⁶ C

q₂ = magnitude of negative charge = 3 x 10⁻⁶ C

r = distance between the two charges = 0.250 m

d = distance of the location of third charge from negative charge

q = magnitude of charge on third charge

Using equilibrium of electric force on third charge

\frac{kq_{2}q}{d^{2}} = \frac{kq_{1}q}{(r+d)^{2}}

\frac{q_{2}}{d^{2}} = \frac{q_{1}}{(r+d)^{2}}

\frac{(5\times 10^{-6})}{(0.250+d)^{2}} = \frac{(3\times 10^{-6})}{(d^{2}}

d = 0.86 m

4 0
3 years ago
If you throw your annoying little sister down the stairs with a force of 60 N and she weighs 30 kg. What’s her acceleration rate
Fynjy0 [20]

Answer:

Her acceleration rate is death.

Explanation:

4 0
3 years ago
Read 2 more answers
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