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vitfil [10]
3 years ago
10

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

ing. If Molly wants to improve her fitness level, what should she do?
Physics
2 answers:
Afina-wow [57]3 years ago
5 0
She should increase the intensity of her exercise?
lakkis [162]3 years ago
3 0
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.
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2 years ago
A boxer hits a punching bag with a force. Describe what happens.
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The punching bag applies the same amount of force to the boxer’s hand.   (C)

5 0
3 years ago
The electric field strength is 20,000 N/C inside a parallel-platecapacitor with a 1.0 mm spacing. An electron is released fromre
creativ13 [48]

Explanation:

It is given that,

Electric field strength, E = 20,000 N/C

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

F=qE

Also, ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\times 20,000}{9.1\times 10^{-31}}

a=3.51\times 10^{15}\ m/s^2

Using third equation of motion as :

v^2-u^2=2ad

v=\sqrt{2ad}

v=\sqrt{2\times 3.51\times 10^{15}\times 0.001}

v = 2649528.2599 m/s

or

v=2.64\times 10^6\ m/s

So, the velocity of the electron when it reaches the positive plate is 2.64\times 10^6\ m/s. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
A thin, 100 g disk with a diameter of 8.0 cm rotates about an axis through its center with 0.15 J of kinetic energy. What is the
butalik [34]

Answer:

v=2.45 m/s

Explanation:

The rotational kinetic energy is calculated with:

K=\frac{I\omega^2}{2}

The moment of inertia of a disk is:

I=\frac{mr^2}{2}

A point on the rim of a disk will have a speed of:

v=r\omega

Putting all together:

v=r\sqrt{\frac{2K}{I}}=r\sqrt{\frac{2(2K)}{mr^2}}=2\sqrt{\frac{K}{m}}

Which for our values is:

v=2\sqrt{\frac{0.15J}{0.1Kg}}=2.45m/s

3 0
4 years ago
Read 2 more answers
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h
Ratling [72]

Answer

given,

time  = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

a_1=\dfrac{1.39}{10}

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

a_3=\dfrac{5.282}{10}

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = \dfrac{0.3892}{0.139}

ratio = 2.8

6 0
4 years ago
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