Jackie rides her bike to school every day and enjoys playing beach volleyball on weekends. She also helps with tasks around the house, like pulling weeds in the garden and sweeping the porch. Which of the following describes a possible benefit of Jackie's physical activity? (1 point) a Jackie may be less likely to commit to a healthy future. b Jackie may have an increased chance of disease. c Jackie may maintain healthy body fat levels. d Jackie may require more assistance performing daily tasks.

Answer:

Place the convex lens on a 'V' stand. . ii) Light a candle and take it far away from the lens along the principal axis. iii) Adjust the screen on the other side of the lens to get a clear image on it. iv) Measure the distance between the 'V' stand and the screen which gives a rough idea of the focal length of the lens.

Answer:

x10 - 6 +(8+6-4)= 10

10m -10x y0

Answer: 20.4 m

Explanation: speed v upward is v = vo -gt.

When maximum height v = 0, vo -gt = 0 and

T = vo / g = 20 m/s / 9.81 m/s² = 2.04 s

Now height y = vot -0.5 gt^2 = 20,4 m

This is possible to solve using

Conservation of energy Ek= Ep

0.5mv² = mgh. And h = v²/2g. Results same

Answer:

not enough

Explanation:

information given

Answer:

Sample Response: In the lab, I discovered the air pressure, air temperature, and relative humidity conditions that create snow, rain, thunderstorms, fog, and clear skies. I learned that low-pressure areas usually experience snowy, rainy, and stormy weather, and that high-pressure areas usually experience either foggy or sunny weather. Knowing how these atmospheric conditions influence weather help weather forecasters predict weather patterns through maps and other information.

(Sorry this was not fast enough )

3 things that happen to our body when we don't eat nutritious meals:

1. It can lead to bad eating habits that may cause obesity or diabetes

2. It can also cause heart disease and cancer

3. it can also make you more tired, and may contribute to stress.

Poor nutrition can contribute to stress, tiredness and our capacity to work, and over time. It can contribute to the risk of developing some illnesses and other health problems such as being overweight or obese.

Hope this helps; have a great day!

When a large number of magnetic domains are aligned, a strong magnet produces more magnetic attraction compared to a weak magnet and non-magnetic material.

Magnetic induction occurs when a magnetic field is used to magnetize another object.

This process occurs when the magnetic domains are aligned in the same direction, which causes the object become magnetized in that direction, thereby forming a magnet.

Thus, when a large number of magnetic domains are aligned, a strong magnet produces more magnetic attraction compared to a weak magnet and non-magnetic material.

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A property of cells, tissues, and organisms that allows the maintenance and regulation of the stability and constancy needed to function properly. Homeostasis is a healthy state that is maintained by the constant adjustment of biochemical and physiological pathways.

have a good day :).

Answer:

4 meters per second

Explanation:

Answer:

If all these three charges are positive with a magnitude of [tex]1.8 \times 10^{-8}\; \rm C[/tex] each, the electric potential at the midpoint of segment [tex]\rm AB[/tex] would be approximately [tex]8.3 \times 10^{3}\; \rm V[/tex].

Explanation:

Convert the unit of the length of each side of this triangle to meters: [tex]10\; \rm cm = 0.10\; \rm m[/tex].

Distance between the midpoint of [tex]\rm AB[/tex] and each of the three charges:

Let [tex]k[/tex] denote Coulomb's constant ([tex]k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].)

Electric potential due to the charge at [tex]\rm A[/tex]: [tex]\displaystyle \frac{k\, q}{d({\rm A})}[/tex].

Electric potential due to the charge at [tex]\rm B[/tex]: [tex]\displaystyle \frac{k\, q}{d({\rm B})}[/tex].

Electric potential due to the charge at [tex]\rm A[/tex]: [tex]\displaystyle \frac{k\, q}{d({\rm C})}[/tex].

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of [tex]\rm AB[/tex] due to all these three charges  would be:

[tex]\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}[/tex].

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = [tex]\epsilon_o \frac{A}{d}[/tex]

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = [tex]\frac{Q}{\Delta V}[/tex]

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = [tex]\frac{Q_o}{ 2C}[/tex]

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

What does "the following" mean ?

Kinetic energy is directly proportional to speed² .

So kinetic energy doubles if the speed is increased by √2 .

(3 m/s) · (√2)  =  4.243 m/s

Answer:

238.09 cm^3

Explanation:

V=mass/density

V=2500/10.5

V=238.09 cm^3

Answer:

The tension on the slack segment of the belt is 65 N

Explanation:

The given parameters are;

The mass of the solid disk = 70.5 kg

The diameter of the flywheel disk, D = 1.25 m

The radius of the pulley = 0.230 m

If the tension in the upper segment of the belt = 165 N

The angular acceleration of the flywheel = 1.67 rad/s²

The moment of inertia of a circular disc, I = 1/2·M·r²

The radius of the flywheel, r = D/2 = (1.25 m)/2 = 0.625 m

The moment of inertia of the flywheel, I = 1/2 × 70.5 kg × (0.625 m)² = 13.7695312 m²·kg

The moment of inertia, I = 13.7695312 m²·kg

Torque, τ = Moment of Inertia, I × Angular acceleration, α

Torque, τ = I × α

τ = 13.7695312 m²·kg × 1.67 rad/s² = 22.9951171 N·m ≈ 23 N·m

The torque applied from the pulley side is given as follows;

Let [tex]T_{slk}[/tex] represent the tension on the slack segment of the belts, we have;

23 N·m = (165 N - [tex]T_{slk}[/tex]) × 0.230 m

∴ 165 N - [tex]T_{slk}[/tex] = 23 N·m/0.230 m = 100 N

[tex]T_{slk}[/tex] = 165 N - 100 N = 65 N

The tension on the slack segment of the belt, [tex]T_{slk}[/tex] = 65 N

The tension on the slack segment of the belt is 65 N

flywheel, heavy wheel attached to a rotating shaft so as to smooth out delivery of power from a motor to a machine. The inertia of the flywheel opposes and moderates fluctuations in the speed of the engine and stores the excess energy for intermittent use.

The given parameters are;

The mass of the solid disk = 70.5 kg

The diameter of the flywheel disk, D = 1.25 m

The radius of the pulley = 0.230 m

If the tension in the upper segment of the belt = 165 N

The angular acceleration of the flywheel = 1.67 rad/s²

The moment of inertia of a circular disc, I = 1/2·M·r²

The radius of the flywheel, r = D/2 = (1.25 m)/2 = 0.625 m

Now we will  calculate tension on the slack side

The moment of inertia of the flywheel,

[tex]I = \dfrac{1}{2} \times 70.5 kg \times (0.625 m)^2 = 13.7695312 \ m^2 \ kg[/tex]

The moment of inertia, I = 13.7695312 m²·kg

Torque, τ = Moment of Inertia, I × Angular acceleration, α

Torque, τ = I × α

τ = [tex]13.7695312 \times 1.67 = 22.9951171 Nm = 23 Nm[/tex]

The torque applied from the pulley side is given as follows;

Let  represent the tension on the slack segment of the belts, we have;

23 N·m = (165 N - ) × 0.230 m

∴ 165 N -  = 23 N·m/0.230 m = 100 N

= 165 N - 100 N = 65 N

The tension on the slack segment of the belt,  = 65 N

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Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

[tex]Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\[/tex]

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

[tex]340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\[/tex]

v₂ = 5.7 m/s

Answer:

5.7 I got it right on edg 2021

Explanation:

Answer:

Hello your question lacks some information but i will give you a more General answer about differences in temperature of bodies on same latitude

answer : Continentality is the reason

Explanation:

Sensors or objects on the same latitude are supposed to have the same air temperatures but in some cases this might not be true. This is because of the proximity of the bodies to an Ocean or any water body.  

Areas close to water bodies experience modest temperatures even if they are on the same latitude with Areas experiencing  harsh temperatures. This effect is termed as Continentality

Answer:

Solution:

Explanation:

Here, mass of body ( m)=12kg.

velocity (v)=37.5 m/s. momentum = mass× velocity = 12kg×37.5m/s =450kgm/s ans.

Answer:

Joules/second^2=(Newton×meter)/second^2.

Solving for mass: Kg= (second^3/ meter^2).

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N[/tex]

Therefore, the force of attraction between the two charges is 3.55 N.

Answer:

knee to chest

Explanation:

I took the test

Answer:

C. Knee to chest.

Line - A:  (15/10)  =  1.5 inch/second

Line - B:  (0/10)  =  0

Line - C:  (10/10)  =  1.0 inch/second

Line - D:  (-25/20)  =  -1.25 inch/second

Line-A represents the greatest speed.

Answer:

The desert biome is an ecosystem that forms due to the low level of rainfall it receives each year. Deserts cover about 20% of the Earth. There are four major types of desert in this biome - hot and dry, semiarid, coastal, and cold. They are all able to inhabit plant and animal life that are able to survive there. Although the daytime temperatures of the desert biome are very hot, they can get very cold at night. The Sahara Desert is the largest desert in the desert biome. It covers over 300 million square miles.

Explanation:

Answer:

Convergent boundaries. When two plates collide and one is pushed over the other, forming a mountain.

Answer:

The Answer is B: to provide traction so that slipping does not occur

Explanation:

if it is wrong please correct me

Answer:

athlete  W = 2560 J

roce      W_{fr} = - 2560 J

Explanation:

For this exercise we can use the relationship between work and kinetic energy

          W = ΔK

          W = K_f -K₀

as the bearing of rest its v₀ = 0 and the final velocity is v = 8 m / s

athlete's job is

           W = ½ m [tex]v_{f}^2[/tex] - 0

let's calculate

           W = ½ 80 8²

            W = 2560 J

For the work of the friction force let us use that since the athlete goes at constant speed, they relate it to zero, therefore the forces are in equilibrium

             ∑F = 0

             F- fr = 0

             F = fr

Therefore, the magnitude of the friction force is equal to the force applied by the athlete, as the friction force always has the opposite direction to the movement, its work is

             W_{fr} = - W_{atlete}

             W_{fr} = - 2560 J

Answer:

No, he hasn't done any work.

Explanation:

In this scenario, the weightlifter hasn't done any work because even though he lifted the bell for a specific period of time, he didn't cover any distance.

Mathematically, work done is given by the formula;

Work done = force * distance

[tex] W = F * d[/tex]

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

[tex] Power = \frac {Energy}{time} [/tex]

Answer:

It will increase.

Explanation:

Because kinetic energy is proportional to the square of the velocity, when velocity is doubled, kinetic energy is quadrupled. Kinetic energy K=mv^2/2

Answer:D. Watching the left-over dirt from a comet burn up by friction as the pieces hit the Earth's atmosphere

Explanation: A meteor shower is giant rocks moving across the sky

There must be an invitation for d. watching the left-over dirt from a comet burn up by friction as the pieces hit the Earth's atmosphere

Meteor showers occur when dust or particles from asteroids or comets enter Earth's atmosphere at very high speed. When they hit the atmosphere, meteors rub against air particles and create friction, heating the meteors. The heat vaporizes most meteors, creating what we call shooting stars

hence correct option will be D) Watching the left-over dirt from a comet burn up by friction as the pieces hit the Earth's atmosphere

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Answer:

Efficiency

Explanation:

Answer:

friction force

Explanation:

friction forces a force that opposed the relative motion of two bodies in contact or when sliding