Help me! Btw he’s playing golf.

Molly is jogging on the treadmill. After 20 minutes, her heart rate has not increased, she is able to talk, and she is not sweat

lakkis [162]

Molly could increase her heart rate by turning around and jogging backwards, by jogging on her hands instead of her feet, or by continuing to jog normally but increasing the speed of the treadmill.

3 0

3 years ago

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A 3.59 kg block is released from rest 53.0 m above the ground. When it has fallen 21.3 m, what is its kinetic energy

Juli2301 [7.4K]

<h3><u>Answer:-</u></h3>

$\implies$ 1115.2 Joules

<h3><u>Given :- </u></h3>

$\red{\leadsto}\:$ $\textsf{Height\: at \:rest\:}$ $\sf,h_2= 53.0m$

$\green{\leadsto}\:$ $\textsf{Height\: to\:which\;block\:}$ $\sf,h_1= 21.3$

<h3><u>Solution</u> :-</h3>

$\maltese$ As the block falls , the change in potential energy will be converted into kinetic energy

$\leadsto$ Potential Energy at rest at height, $h_2$ = $\sf mgh_2$

$\leadsto$ Potential Energy when fell at height, $h_1$ = $\sf mgh_1$

$\\$

$\leadsto$ Kinetic energy= Change in Potential energy = = $\sf mgh_2-mgh_1$

$\begin{gathered}\implies\quad \sf mg(h_2-h_1) \\\end{gathered}$

$\begin{gathered}\implies\quad \sf mg(53.0-21.3) \\\end{gathered}$

$\begin{gathered}\implies\quad \sf 3.59 \times 9.8\times (53.0-21.3) \\\end{gathered}$

$\begin{gathered}\implies\quad \sf 3.59 \times 9.8\times 31.7\\\end{gathered}$

$\begin{gathered}\implies\quad \boxed{\sf{1115.2\: Joules}}\\\end{gathered}$

7 0

3 years ago

A volume of air increases 0.227 m^3 at a net pressure of 2.07 x 10^7 Pa. How much work is done on the air?

aliya0001 [1]

Answer:

The work done on the air is 4.699 x 10⁶ Joules

Explanation:

Given;

increase in air volume, ΔV = 0.227 m³

net pressure of the air, P = 2.07 x 10⁷ Pa

The work done on the air is given by;

W = PΔV

Where;

W is the work done on the air

P is the net pressure

ΔV is the increase in air volume

Substitute the given values and solve for work done;

W = (2.07 x 10⁷ Pa) (0.227 m³)

W = 4.699 x 10⁶ Joules

Therefore, the work done on the air is 4.699 x 10⁶ Joules

7 0

3 years ago

An automobile traveling 92.0 km/h has tires of 64.0 cm diameter. (a) What is the angular speed of the tires about their axles? (

Semenov [28]

Answer:

76.4035 m

Explanation:

r = Radius = 0.32 m

$\omega_f$ = Final angular velocity = 0

$\omega_i$ = Initial angular velocity = 92 km/h

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{\dfrac{92}{3.6}}{0.32}\\\Rightarrow \omega=79.861\ rad/s$

The angular speed of the tires about their axles is 79.861 rad/s.

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-79.861^2}{2\times 2\pi \times 38}\\\Rightarrow \alpha=-13.355\ rad/s^2$