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3 years ago

PLEASE ANSWER ASAP!!! ITS DUE IN 10 MINUTES!!!!!!

Physics

1 answer:

forsale [732]3 years ago

5 0

Explanation:

2. Aerobic exercise. ...

Strength building. ...

Balance Training. ...

Endurance. ...

Flexibility. ...

Moderate intensity exercise. ...

Vigorous exercise.

3.Brisk walking.

Running.

Jogging

Bear crawls.

Swimming.

Water aerobics.

Cycling/bicycling.

4. Running or Jogging. ...

Outdoor Cycling. ...

Walking.

5. The type of exercise to build muscle is strength training, although cardiovascular activity can also provide benefits.

Hope it helps..

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The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

jek_recluse [69]

Answer:

$e=3367.2J$

% $e=1.43$ %

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

$v_{r}=10km/h=2.77m/s$

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

$v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s$

Knowing that we can calculate Kinetic energy for the real and experimental speed

$E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J$

$E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J$

Now, the potential error in her calculated kinetic energy is:

$e=E_{r}-E_{e}=(234023-230656)J=3367.2J$

% $e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43$ %

4 0

3 years ago

Partially correct answer iconYour answer is partially correct. A proton initially has and then 2.30 s later has (in meters per s

morpeh [17]

Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.

a) = (1.82i - 9.69k)m/s2

b) magnitude = 9.85m/s2

c) direction = 280.64 degree

Explanation:

The detailed and step is shown in the attachment.

7 0

3 years ago

When a very cold air mass covers half of the united states, a very warm air mass often covers the other half. explain how this h

Rudiy27

Maritime tropical air masses develop over warm waters present in the tropics and Gulf of Mexico, where heat and moisture are carried to to the overlying air from the water below.

 Tropical air masses having northward movement carry warm moist air into the United States, thus increasing the potential for condensation. Generally the southern states experience tropical air masses. But, in winter season, southerly winds ahead of migrating cyclones sometimes transport tropical air mass towards north.

The counterclockwise winds related to northern hemisphere mid latitude cyclones play an important role in the movement air masses, carrying warm moist air towards north ahead of a low while dragging colder and drier air towards south.

4 0

3 years ago

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$