10. when creating a cardiovascular fitness routine which program would be appropriate?
A Sprint a short distance
B Lift a heavy weights
C Run for a long time
D Use the entire range of motion

13. what would be an appropriate amount of time to hold a static stretch?
A 5 seconds
B 20 seconds
C 1 minute
D 5 minute

14. what activity may help prevent injury if it is performed prior to playing a sport?
A weight thing
B dynamic stretching
C high intensity interval training
D sprints

15. when your workout routines gets too easy what should you do?
A Increase the frequency
B Increase the intensity
C Increase the time
D All​




PLEASE HELP

A) The rover’s coordinates and its distance from the lander at t = 2.0 s are; (1, 4) and 4.1 m

B) The rover’s displacement and average velocity vector during the interval are; s = (-1, 4) and v = (-0.5, 2) m/s

C) The magnitude and direction of the instantaneous velocity are; 2.24 m/s and 117°

The rover's x and y coordinates are given as;

x = 2.0m − (0.25 m/s²)t²

y = (1.0m/s)t + (0.25 m/s³)t³

A) At t = 2 s, the rovers coordinates are;

x = 2.0m − (0.25 m/s²)2²

x = 1 m

y = (1.0m/s)2 + (0.25 m/s³)2³

y = 4 m

Distance from the lander is;

s = √[(1 - 2)² + 4²]

s = 4.1 m

B) Let us first find the distance coordinates for the interval t = 0.0 s to t = 2.0s. Thus;

s = r - r₀

s = (1 - 2), (4 - 0)

s = (-1, 4)

Thus, average velocity vector is;

v = ¹/₂s

v =  ¹/₂(-1, 4)

v = (-0.5, 2) m/s

C) A general expression for the instantaneous velocity components is;

v_x = -0.5t

v_y = 1 - 0.75t²

Thus, v(2) is;

v_x = -0.5(2) = -1

v_y = 1 - 0.75(2)²

v_y = -2

Instantaneous velocity vector is; v = (-1, -2)

Magnitude of instantaneous Velocity = √(-1² + -2²) = 2.24 m/s

Direction = 180° - tan⁻¹(-2/-1) ≈ 117°

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Answer:

-240

Explanation:

A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.

The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.

Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.

When the body slows down from its present level to a condition of rest, the same quantity of energy is used.

Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.

As per the given information in the question,

Distance, d = 2.0 m

Friction, f = 120 N

The angle between displacement and friction force, θ = 180°

Now, the change in kinetic energy for the motorcycle = Work done by the friction.

K.E = f × d(cos θ)

= 120 (2.0 m)(cos 180°)

Δ K.E = -240 J

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The left hemisphere of the brain is primarily needed in scenario, Solving a complex mathematical problem. Option a is correct.

The brain is divided into two hemispheres, left and right, and they are specialized for different cognitive functions. The left hemisphere of the brain is primarily responsible for language processing, logical reasoning, and analytical thinking. Solving a complex mathematical problem involves logical reasoning, analytical thinking, and the use of language, all of which are primarily controlled by the left hemisphere of the brain.

Mathematical problems often require precise calculations, sequencing of steps, and the use of symbols and formulas, all of which require a strong left-brain function. In contrast, appreciating a work of art, listening to music, and recognizing facial expressions are all more complex perceptual and emotional processes that involve the right hemisphere of the brain. Option a is correct.

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--The complete question is, In which of the following scenarios is the left hemisphere of the brain primarily needed?

a. Solving a complex mathematical problem.

b. Appreciating a work of art

c. Listening to music

d. Recognizing facial expressions.--

Answer:

I. 6 cells .

II. Series connection.

Explanation:

I. Determination of the number of cells needed.

From the question given above,

Total voltage (V) = 9 V

1.5 V = 1 cell

Number of cells needed =?

The number of cells needed to make the 9V battery can be obtained as follow:

1.5 V = 1 cell

Therefore,

9 V = 9 V × 1 cell / 1.5 V

9 V = 6 cells

Thus, 6 cells of 1.5 V each is needed

II. Determination of the connection line

Total voltage (Vₜ) = 9 V

Cell 1 (V₁) = 1.5 V

Cell 2 (V₂) = 1.5 V

Cell 3 (V₃) = 1.5 V

Cell 4 (V₄) = 1.5 V

Cell 5 (V₅) = 1.5 V

Cell 6(V₆ ) = 1.5 V

For parrall connection:

Vₜ = V₁ = V₂ = V₃ = V₄ = V₅ = V₆

9 V = 1.5 V =... = 1.5 V

For series connection:

Vₜ = V₁ + V₂ + V₃ + V₄ + V₅ + V₆

9 = 1.5 + 1.5 + 1.5 + 1.5 + 1.5 + 1.5

9 V = 9 V

From the illustration above, we can see that series connection of each cells will give a total volt of 9 V unlike the parallel connection which resulted to 1.5 V.

Therfore, the cells should be arranged in series connection

Answer:

  C.  98 J

Explanation:

The appropriate formula is ...

  PE = mgh . . . . . m is mass; below, m is meters

  PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2

  PE = 98 J

Answer:

Focal length (f) = 1/2 (Object distance + Image distance)

f = 1/2 (12.2 cm + 25.0 cm)

f = 18.6 cm

Answer:

er so ue 0.984 kg study your self brhh

Answer:

Telescopes have changed the way scientists study the Solar System. Before telescopes, people were only able to observe the Universe with their naked eyes. ... If it were not for telescopes, scientists would not be able to detect possible threats to Earth or predict comet's sightings.

Explanation:

400 years ago, before telescopes, our understanding of the universe was very different.

This is what was believed:

We live on a spherical ball orbited by the rest of a finite, spherical universe.

Earth does not move. It is the center of the universe.

Our Sun orbits the Earth, as do all the other planets and the Moon.

The stars are distant objects, always perfect and unchanging.

How did telescopes & associated technologies unlock the secrets of the universe and

help us toward the understanding we have today where Earth is no longer at the center of

the universe? Instead, we know that ours is a small planet orbiting a star in the suburbs

of a large galaxy filled with billions of other stars and planets, surrounded by billions of

other galaxies becoming increasingly ever distant from each other by the expansion of

space.

This is the story of how telescopes continuously changed our understanding of the

universe and our place in it - transforming our view of our universe. And we still have

much more to discover!

2. Before telescopes, we could only use our eyes and a variety of measuring instruments to

plot the positions and movements of objects in the sky to create a limited understanding

of our universe. We had no way to know what these objects actually were and little

evidence for our relationship to the cosmos.

1) Given,

The rate is $0.10 per kilowatt hour (KWh) with a consumption of 900

KWh.

Solution:

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer:

You would weigh the same.

Explanation:

At the moment, since Earth is not a perfect sphere, the Earth "bulges out" at the equator, so you're further from the centre of the Earth. Since gravity acts through a body's center of mass, the further you are from the centre the weaker the gravitational acceleration you will feel, because gravity weakens over distance.

So, you're actually lighter at the equator than you'd be at the poles.

However, if the Earth was a perfect sphere, this "bulge" at the equator would not happen, and so you would weigh the same at the poles and at the equator.

Hope this makes sense.

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Time of flight = 1.6 s

Horizontal distance = 64 m

Projectile motion is the form of motion experienced by an object or particle projected into a gravitational field, such as from the surface of the Earth, and moves along a curvilinear path only under the action of gravity.

For the given case,

h = vt + ¹/₂gt²

h = height of tower

v = initial velocity

t = time of flight

55 = 50sin30t + ¹/₂9.8t²

55 = 25t + 4.9t²

4.9t² + 25t - 55 = 0

t = 1.6 s

X = vₓt

X = horizontal distance

vₓ = horizontal velocity

t = time of flight

X =  (50 x cos30) x 1.6

X =  64 m

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Use a digital multimeter and disconnect the negative battery terminal when measuring resistance on a CANbus.

A multimeter, also called  volt-ohm-meter ( VOM ),is a handheld measuring instrument either analog or digital that can measure the  voltage, resistance, and current across circuits in a device.

When measuring resistance on a CANbus with the digital multimeter, it is  necessary to disconnect the negative  battery , This puts the power off. This is to ensure only the resistance is measured and prevent  other components'  from being measured and distorting readings.

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The height of the acorn above the ground before falling is 2.9 m.

The initial velocity of the acorn is calculated as follows;

s = ut + ¹/₂gt²

where;

1.87 = 0.311u   +   ¹/₂(9.8)(0.311²)

1.87 = 0.311u  +  0.474

0.311u = 1.396

u = ( 1.396 ) / ( 0.311 )

u = 4.489 m/s

The height the acorn falls before reaching the top velocity of 4.489 m/s is calculated as follows;

v² = u² + 2gh

where;

v² = 0 + 2gh

h = v² / 2g

h = ( 4.489² ) / ( 2 x 9.8 )

h = 1.03 m

The total height of the oak tree = 1.87 m + 1.03 m = 2.9 m

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A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.

A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.

A positron is an electron with a positive charge.

During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.

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Complete Question

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 32.75 m/s at an angle θ = 32° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.

Randomized Variables

vo 32.75 m/s

theta 32 degrees

What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?

Answer:

The  horizontal distance is  \(R =  590.2 \ m \)  

Explanation:

From the question we are told that

 The initial  velocity is  \(v_o =  32.75 \  m/s\)

  The angle is  \(\theta =  26^o\)

  The gravitational acceleration of the moon is \(g_m  =  \frac{1}{6}  *  9.8   = 1.633 m/s^2\)

 Generally the distance traveled is mathematically represented as

    \(R =  \frac{v_o^2 sin 2(\theta)}{g_m}\)

=> \(R =  \frac{32.75^2 sin 2(32)}{1.633}\)

=> \(R =  590.2 \ m \)  

Answer:

Solids - Bricks , wood , Pottery, Bucket

Liquid - Water, soap, Sanitizers.

Gases - Aerosol in Deodorants, Chlorofluorocarbons in Fire extinguishers , Butane in lighters.

Answer:

0.0512 kJ

Explanation:

From the given information, since the values were not given:

Then, Let assume that:

A set of 50 titanium rods is handled by the engineer, with 50cm length and 20g of mass.

The overhead walkway = 5.3 m

In this process, 12 rods roll of the walkway

We are to determine the change in the Gravitational P.E

Change in P.E = final P.E- initial P.E

Change in P.E = (50 mgh) - (0)

Change in P.E =  \((50\times \dfrac{20}{1000}\times 9.8 \times 5.3 )-0\)

Change in P.E =  51.94 Joules

Change in P.E = 0.05194 Joules

Change in P.E ≅ 0.0512 kJ

Answer:

The correct answer is C) Force. If the total force on an object is not zero, its motion will change according to the second law of motion by Isaac Newton.

Answer:

First blank: 251 N

Second blank: 691 N

Explanation:

Answer:

B is the correct answer

Explanation:

I just did this question

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

We are given that an acorn is about to fall from a tree. This means that the acorn is not accelerated yet and is at a height "h" from the ground. The energy associated to ab object at a certain height is:

Where:

The energy "U" is called Gravitational Potential Energy (GPE).

The height that will illustrate the distance will be d = 6.36m

Based on the information given, the length of the vine will be:

L = ✓(16² + 27.7)²

L = 32m

The velocity of Jane when she reaches position B will be:

V = ✓2gh

V = ✓(2 × 9.8 × 4.3)

V = 9.18m/s

We will apply the conversation of momentum. This will be:

50 × 9.18 = (50 + 80)V1

V1 = 3.53m/s

Therefore, the height that will illustrate the distance will be:

31.36² + d² = 32²

d² = 32² - 31.36²

d = 6.36m

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Answer:

The correct answer is Dean has a period greater than San

Explanation:

Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.

                T² = (4π² / G M)  r³

When applying this equation to our case, the planet with a greater orbit must have a greater period.

Consequently Dean must have a period greater than San which has the smallest orbit

The correct answer is Dean has a period greater than San

Answer:

According to the law of universal gravitation, any two objects are attracted to each other. The strength of the gravitational force depends on the masses of the objects and their distance from each other.

Many stars have planets around them. If there were no gravity attracting a planet to its star, the planet's motion would carry it away from the star. However, when this motion is balanced by the gravitational attraction to the star, the planet orbits the star.

Two solar systems each have a planet the same distance from the star. The planets have the same mass, but Planet A orbits a more massive star than Planet B.

Which of the following statements is true about the planets?

 A.

Planet B will keep orbiting its star longer than Planet A.

 B.

Planet A has a longer year than Planet B.

 C.

Planet A orbits its star faster than Planet B.

 D.

Planet B is more attracted to its star than Planet A.

Explanation: