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She would be better off fitting in 15-20 minutes of exercise several times a week. This t is the BEST advice for Evonne.

Regular physical activity and exercise have many positive health effects that are difficult to deny. Everyone, regardless of age, sex, or physical ability, benefits from exercise.

Exercise can assist sustain weight loss or prevent excessive weight gain. Calorie burn occurs during physical exertion. You burn more calories when you engage in more vigorous exercise.

Numerous health issues and concerns, such as stroke, the metabolic syndrome, high blood pressure, and type 2 diabetes, can be prevented or managed with regular exercise. It can also help with cognitive improvement and reduce the danger of dying from any cause.

Exercise helps your circulatory system function more effectively and distributes oxygen and nutrients to your tissues. Additionally, you have greater energy to complete daily tasks as your heart and lung health improves.

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Gas.

HOPE YOU GET 100!

Answer:

750 Newton

Explanation:

force=mass(acceleration)

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

[tex]\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\[/tex]

To get the required work done, we will divide the mass by the speed of one knot to have:

[tex]w=\frac{7230}{0.51}\\w= 14,176.47Joules[/tex]

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

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The minimum force required to start this block moving is 370 N.

The given parameters;

The minimum force required to start this block moving is calculated as follows;

[tex]F= \mu_s F_n[/tex]

where;

[tex]F_n[/tex] is the normal force on the block which is equal to the weight of the block

[tex]F= \mu_s F_n \\\\F= \mu_s W\\\\F = 0.74 \times 500 \\\\F= 370 \ N[/tex]

Thus, the minimum force required to start this block moving is 370 N.

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Answer:

15 N :D

Explanation:

The size of the drag force on this object falling at a constant speed is determined as 15 N.

The drag force on the object is determined by calculating the net force on the object;

F(net) = W - D

where;

ma = W - D

at constant speed, acceleration, a = 0

0 = W - D

W  = D

15 N = D

Thus, the size of the drag force on this object falling at a constant speed is determined as 15 N.

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Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

Answer:

It could be low self-esteem

Explanation:

There probably insecure have problems and issues of their own could be that were abused growing up and weren’t treated right who knows.

Answer:

90 Km/h

Explanation:

60 mins is 1 hr

so in an hour bus covers 60 Km

so new speed:

d/t

(60km/40mins)*60mins/h

90Km/h

Answer:

The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.

The wave height is equal to twice the amplitude of the wave.

The wave height of a wave of given wave with amplitude, period and wavelength is equal to twice the amplitude of the wave.

The amplitude of a wave is the maximum displacement of the wave, starting from the zero position of the wave. The wave height measures twice the maximum displacement of the wave.

Thus, we can conclude that the wave height is equal to twice the amplitude of the wave.

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Answer:

yes

Explanation:

because it is a diverging mirror

Answer:

hi I don't know sorry sorry forgive me

Explanation:

sorry

Answer: if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.

Explanation:

Answer:

There must be a lot of dark matter that can be felt but not seen

The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.

Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.

A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.

The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.

Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.

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Gravitational Potential Energy of an object is calculated by formula ~

[tex] \large\boxed{\sf P = mgh}[/tex]

where,

Now, let's calculate its potential energy ~

Answer:

Potential Energy of an object is calculated by formula:

Where,

Now, let's solve the question.

Given,

Now,

We know that,

Potential Energy (P.E)=mgh

[tex] = 3500 \times 10 \times 4[/tex]

[tex] = 3500 0\times 4[/tex]

[tex] =140000 joules [/tex]

[tex]\mathfrak{\blue{DisneyPrincess29}}[/tex]

The work done on the given gas in one cycle is -405.3 kJ.

The given parameters;

Convert the pressure to Pascal (N/m²);

1 atm = 101325

The work done in one cycle is the area of the triangle and it is calculated as follows

[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]

the net work done on the gas = - 405.3 kJ

Thus, the work done on the given gas in one cycle is -405.3 kJ.

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Answer:

55.84 look at periodic table

so 55.84*3= 167.52gFe

Explanation:

300 second = 5 minutes

details

500 watts = 500 joules per second

Time = (1.50 x 10^5 joules)/(5 x 10^2 joules per second) = .3 x 10^3 seconds =300 second

Explanation:

the figure in the left side has higher frequency.

because it has more nos. of wave in 1sec.

Answer:

1.84 kJ  (kilojoules)

Explanation:

A specific heat of 0.46 J/g Cº means that it takes 0.46 Joules of energy to raise the temperature of 1 gram of iron by 1 Cº.

If we want to heat 50 g of iron from 20° C to 100° C, we can make the following calculation:

Heat = (specific heat)*(mass)*(temp change)

Heat = (0.46 J/g Cº)*(50g)*(100° C -  20° C)

[Note how the units cancel to yield just Joules]

Heat = 1840 Joules, or 1.84 kJ

[Note that the number is positive:  Energy is added to the system.  If we used cold iron to cool 50g of 100° C water, the temperature change would be (Final - Initial) or (20° C - 100° C).  The number is -1.84 kJ:  the negative means heat was removed from the system (the iron).

The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.

We have to find the acceleration using;

v = u - at

v = final velocity = 0 m/s

u = initial velocity = 35.6m/s

a = acceleration = ?

t = time = 0.35 s

u = at

a = u/t = 35.6m/s / 0.35 s

a = 101.7 ms-2

The force on the driver =  75kg ×  101.7 ms-2 = 7629 N

Using the principle of conservation of momentum;

Momentum before collision = momentum after collision

m1u1 +m2u2 = m1v1 + m2v2

Hence

(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2

13.12 = 4.8 + 2.3v2

13.12 - 4.8 = 2.3v2

v2 = 13.12 - 4.8/2.3

v2 = 3.6 m/s

Using the principle of conservation of linear momentum;

m1u1 + m2u2 = m1v1 + m2v2

(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v

v = 79.98 + 55.1/10.1

v = 13.37 m/s

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The maximum displacement angle of the bob is 13⁰.

The given parameters;

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]

The maximum displacement angle is calculated as follows;

[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]

Thus, the maximum displacement angle of the bob is 13⁰.

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Answer:

The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)

Explanation:

Answer:

The greater the amplitude of a wave then the more energy it is carrying.

Explanation:

E is carrying the most energy which means that it has the highest amplitude

A has the least energy because it has the lowest amplitude

The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

Pressure is defined by the relationship between force and area.

       P = F / A

The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.

          PV = nR T

Where P is pressure, V is volume, and T is temperature.

Let's write this equation for two points assuming that the temperature has not changed.

          P₀ V₀ = P₁ V₁

          V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex]                 (1)

The subscript "o" is used for the start point and the subscript "1" for the end point.

The pressure in a barometer is:

         P = ρ g y

They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.

The volume of the cylinder is

         V = π r² y

Let's calculate the initial volume.

         V₀ = π 1 9

         V₀ = 28.27 cm³

We substitute in equation 1.

         V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]  

         V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]  

Let's calculate.

        V₁ = [tex]\frac{75}{59} \ 27.27[/tex]  

        V₁ = 35.94 cm³

The volume to be incremented is

         ΔV = V₁ - V₀

         ΔV = 35.94 - 28.27

         ΔV = 7.67 cm³

Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

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Answer:

PLEASE MAARK ME BRAINLEST!!!

Explanation:

The particles of the medium just vibrate in place. As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on.

Answer:

[tex]8.0\; \rm J[/tex].

Explanation:

The efficiency of a machine is the ratio between the useful output and the energy input:

[tex]\begin{aligned}\text{efficiency} &= \frac{\text{useful output}}{\text{energy input}} \times 100\%\end{aligned}[/tex]

Rearrange this equation to find energy input in terms of efficiency and useful output:

[tex]\displaystyle \text{energy input} = \frac{\text{useful output}}{\text{efficiency} / (100\%)}[/tex].

Substitute in the values: [tex]\text{useful output} = 1.2\; \rm J[/tex] and [tex]\text{efficiency} = 15\%[/tex]. Evaluate to find the value of [tex]\text{energy input}[/tex]:

[tex]\begin{aligned} \text{energy input} &= \frac{\text{useful output}}{\text{efficiency} / (100\%)} \\ &= \frac{1.20\; \rm J}{15\% / (100\%)} \\ &= 8.0\; \rm J\end{aligned}[/tex].

(Rounded to two significant figures as in the value of efficiency.)

Hi there!

We can go about this problem using a summation of torques.

In order to ensure the rod is in equilibrium, we must satisfy the condition:

Στ = 0

Since the rod is "very light", we can disregard its mass.

The equation for torque is:

τ = rFsinθ

In this instance, the torques are the weights of the objects and their distance from the pivot.

As the rod is 40 cm, the pivot is at 20 cm. Also, the torques must sum up to 0, so:

0 = rF1 - rF2

r1F1 = r2F2

0.20(50) = r2(200)

Solve:

10 = r2(200)

r2 = 0.05 m = 5 cm

The 200N weight must be put at a distance of 5 cm from the OTHER SIDE of the pivot in order to balance the rod.