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Debora [2.8K]
3 years ago
8

A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at

this instant?
KEi = _________________J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=______________
Physics
1 answer:
Liono4ka [1.6K]3 years ago
4 0

Answer:

(a)  the runner's kinetic energy at the given instant is 308 J

(b)  the kinetic energy increased by a factor of 4.

Explanation:

Given;

mass of the runner, m = 64.1 kg

speed of the runner, u = 3.10 m/s

(a) the kinetic energy of the runner at this instant is calculated as;

K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2}  \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J

(b) when the runner doubles his speed, his final kinetic energy is calculated as;

K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J

the change in the kinetic energy is calculated as;

\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4

Thus, the kinetic energy increased by a factor of 4.

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Read 2 more answers
Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

4 0
3 years ago
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

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Answer:

a) the angular acceleration is 80 rad/s²

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- a = 48 m/s²

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Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

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Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

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Answer:

it is ---D because you can't measure gas and it's mass

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