Across:
5. The term used for describing the
bending of an arm or a leg
6.
9.
Down:
1. This is one popular brand of exercise
machine
2.
They are made of two types of fibers
You can achieve your target pulse rate
by varying the number of these
flexes the arms at
10. The arm
the elbows
11. These muscle fibers do not require as
much oxygen
13. He wrote a book on aerobics
14. This type of training works primarily
with the white muscle fibers
15.
extends the legs at the
The leg
knees
16. Aerobic exercises help the body's
ability to use this element
18.
The term used to describe the straight-
ening of your arm or leg
3.
4.
5.
6.
7.
8.
This was the most popular form of aerobic
exercise in the early years of aerobics
This organ is one part of the cardiovascu-
lar system
This exercise conditions the cardiovascu-
lar system
Aerobic exercise became part of this
movement in the 1970s
12.
17.
This accompanies aerobic dance
A specific exercise produces this type of
result
This type of training involves moving
from one exercise machine to another with
little rest
If you just train for strength, you will not
improve this
These muscle fibers contract more slowly

The value of the electric current for the given conditions willl be  1939.02 A.

Ohm's law claims that the voltage across a conductor is directly proportional to the current flowing through it.

Ohm's law claims that the voltage across a conductor is direct to the current flowing through it. This current-voltage connection may be expressed mathematically as,

V=IR

15.9 V = I × 8.2 × 10³ Ω

I = 1939.02 A

Hence, the value of the electric current for the given conditions willl be  1939.02 A.

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(a)  The charge stored in the 5.00-μF capacitor is 37.2  μC.

(b) The potential difference across the 10.0-μF capacitor is 3.72 V.

The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.

The capacitance of a capacitor is denoted by C and expressed as

C = Q/V

Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.

(a) The equivalent capacitance is

1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃

Substitute the values, we get

Ceq = 3.1  μF

The charge stored in 5.00-μF capacitor is

Q  = Ceq x V

Q = 3.1  μF x 12 V

Q = 37.2  μC

Thus, the charge stored in the 5.00-μF capacitor is 37.2  μC

(b) The potential difference across the 10.0-μF capacitor is given by

V = Q/C₂

Put the values, we get

V = 37.2 / 10

V = 3.72 V

Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.

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Answer:

the 1 gram wasp

Explanation:

To start off with this problem, write down every piece of information and do neccessary conversions.

mass of bee = 2 grams = 0.002 kg

speed of bee = 1 m/s

mass of wasp = 1 gram = 0.001 kg

speed of wasp = 2 m/s

now, we will use the kinetic energy formula and compare the answers

KE BEE = 0.5 (0.002 kg)(1 m/s)^2 = 0.001 Joules

KE WASP = 0.5(0.001 kg)(2 m/s)^2 = 0.002 Joules

0.002 J > 0.001 J

At 11.3°C the rod will just come into contact

Coefficient of linear expansion of aluminium [tex]\alpha_{Al}[/tex] = [tex]23*10^-^6[/tex] °[tex]C[/tex]

Coefficient of linear expansion of Brass [tex]\alpha_{B} =19*10^-^6[/tex] °[tex]C[/tex]

[tex]\alpha = \Delta{L}/Lo(T2-T1)[/tex]

For aluminium

[tex]\alpha_{Al} = \Delta{L}/Lo(T-0)[/tex]

[tex]\Delta{L_{1}} = (23*10^-^6*50*T)[/tex]

For Brass

[tex]\alphaB = \Delta{L_{2}/Lo(T-0)[/tex]

[tex]\Delta{L_{1} +\Delta{L_{2} =0.024cm(23*10^-^6*50*T)+(19*10-6*50*T) =0.024[/tex]

T =11.43 °C

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Answer:

i think the answer is 20.......

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.

When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.

When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.

So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

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Answer:

Their common velocity after the collision will be 5.5m/s

Explanation:

look at the attachment above ☝️

The distance from the rounded tip of the plate to the center of mass is 1.87 m.

The center of mass is a point inside or outside the mass where all of the mass is concentrated.

The y-coordinate of the centroid is given by the ratio of two definite integrals;

Yc = ∫ydm/∫dm,

where dm is a density function

For the uniform plate, δ does not change with position in the plate.

Yc = ∫yδdA/∫δdA

Yc = ∫ydA/∫dA.

dA is a horizontal slice of the plate with dimensions xdy.

Solving the parabola for x,

y = 0.5x²

x = ± √(y/0.50), where the negative value corresponds to the left half of the parabola and the positive to the right half.

dA = (√(y/0.50)

     = √(y/0.50))dy

     = 2(√(y/0.50))dy

The limits of integration are from zero to 1.870, the top of the plate.

∫ydA = ∫2y√(y/0.50)dy = 7.232 m³

∫dA = ∫2√(y/0.50)dy = 3.868 m²

∫ydA/∫dA =  7.232 m³/3.868 m²

∫ydA/∫dA =  1.869700 m

∫ydA/∫dA =   1.87 m

Thus, the distance from the rounded tip of the plate to the center of mass is 1.87 m.

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This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the speed is equal to 4.6 m/s, the time is equal to 10 s, which is left as follows:

[tex]\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }[/tex]

[tex]\bf{d=46 \ m}[/tex]

Therefore, the speed at 10 seconds is 46 meters.

[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Pisces04}}}}}}[/tex]

The steel would expand by 4. 8 * 10^-3 cm

The change in length  ΔL is proportional to length  L. It is dependent on the temperature, substance, and length.

Using the formula:

ΔL= α LΔT

where  ΔL  is the change in length  L = 10cm

ΔT  is the change in temperature = 60° - 20° = 40° C

 α  is the coefficient of linear expansion = 1.2 x 10^-5 °C

Substitute into the formula

ΔL = [tex]1.2 * 10^-5 * 10 * 40[/tex]

ΔL = [tex]4.8 * 10 ^-3[/tex] cm

Therefore, the steel would expand by 4. 8 * 10^-3 cm

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The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

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Greek terms psyche, which means "soul" or "mind," and logos, which means "the study of," are combined to form the english word psychology. The assertion is true.

Psychology is the name given to the scientific study of the mind and behavior. Psychologists are actively interested in researching and comprehending how the mind, the brain, and behavior work.

Psychology is a combination of the Greek words psyche, which means “soul” or “mind,” and logos, which means “the study of.”The given statement is corect.

Hence the best answer from the choices provided will be true.

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The distance of the Image will be -33.75 cm

A concave mirror has an inward-curving reflecting surface that faces away from the light source. Unlike convex mirrors, a concave mirror's image forms a variety of images based on the object's proximity to the mirror.

Given that, an object placed 27 cm from a concave mirror having the focal length of 15 cm

We have to find distance of the Image

Using Mirror Formula:

1/f = 1/v + 1/u

Where,

f = focal length

v =  Image distance from the mirror

u = object distance from the mirror (concave)

Substitute the known values in the above formula to find the value of 'v' i.e. from the mirror.

1/(-15) = 1/v + 1/(-27)

1/(-15) = 1/v - (1/27)

1/v = -0.029

v = -33.75 cm

Therefore the distance of the Image will be -33.75 cm

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Wave I stands for a wave with an equal frequency as wave II. Option c is correct.

Frequency is defined as the number of repititions of a wave occurring waves in 1 second. Its unit is Hz.

Frequency is given by the formula as,

[tex]\rm f = \frac{1}{t}[/tex]

Where,

f is the frequency

t is the period of the wave

From the digrame it is observed that both the wave has the same period. So that they will have the same frequency.

Hence option c is correct.

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Option C. The measurement with the most significant number is 6.600 cm.

Significant numbers are numbers that have significance or meaning and give more precise details about the value of the entire numbers.

Thus, the measurement with the most significant number is 6.600 cm.

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Answer:

12.9m/s^2

Explanation:

As, a=(v^2)/r

=(58^3)/260

=12.9

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

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The mass and tension due to the system are as follows:

Tension is a type of pulling force due transmitted by means of a string or cable.

a) The mass of the child and scale = 104/9.81 = 10.6 kg

b) The tension T, in the cord attaching the child to the scale = scale reading = 104N

c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale

T' = 104 + (0.5 * 9.81)

T' = 108.9 N

d) The sketch is attached in the picture

In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.

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Answer:

Explanation:

Hope I give all correct answer please mark as brainlest answer

Answer:

A

Explanation:

As you go deeper, the pressure increases because the water is going farther away from the air and atmosphere

Answer:

r= 1.09×10^-4

Explanation:

Given

V=speed=4.0×10^5 m/s

B= magnetic field= 0.040 T

©=angle= 35°

m= mass of electron= 9.11×10^-31

q= charge of electron= 1.60×10^-19

solution

qv×B= mv²/r

qvBsin©=mv²/r

qBsin©=mv/r

r=mv/qBsin©

r=9.11×10^-31× 4.0×10^5/1.064×10^-19×0.04T(sin35°)

r= 1.09×10^-4 m

a) r = 1.09 * [tex]10^{-4}[/tex] m

b) Distance travelled : 6.845 *  [tex]10^{-4}[/tex] m    

An electron is a  stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids.

given

charge of electron : 1.6 * [tex]10^{-19}[/tex] C

mass of electron = 9.11 * [tex]10^{-31}[/tex] kg

v = 4.0 x 10 m/s

B =  0.040 T

theta = 35 degrees

since ,

force in magnetic field on electron = centripetal force

a) q(v*B) = m [tex]v^{2}[/tex] / r

q v B sin(theta) =  m [tex]v^{2}[/tex] / r

r =m v /q B sin(theta)

r =  9.11 * [tex]10^{-31}[/tex] * 4.0 x [tex]10^{5}[/tex]/ 1.6 * [tex]10^{-19}[/tex] sin (35)

r = 1.09 * [tex]10^{-4}[/tex] m

b)  far forward will the electron have moved after completing one circle will be equal to circumference of the circle = 2πr

 = 2 * 3.14 * 1.09 * [tex]10^{-4}[/tex] m  = 6.845 *  [tex]10^{-4}[/tex] m  

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The impulse given to the ball by the floor is 0.2865 kg.m/s.

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

A ball of mass 500g is dropped from a height 1.5m . It rebounds the floor and reached the height 1.2m.

The initial velocity u = √2x 9.81 x 1.5 = 5.425 m/s

The final velocity v = √2x 9.81 x 1.2 = 4.852 m/s

Substitute the values into the expression, we get

Impulse = m(v- u)

Impulse=0.5 x (4.852- 5.425 )

Impulse = - 0.2865 kg.m/s

Thus, the magnitude of impulse given to the ball by the floor is 0.2865 kg.m/s.

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Answer:horse

Explanation:

The speed of the first car just before the collision is 26.73 km/h.

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given a 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up and slide together with a speed of 15.0 km/h.

Second car is parked, so its velocity will be zero.

808 x u +632 x 0 = (808 +632 ) x 15

u  = 26.73 km/h

Thus, the speed of the first car just before the collision is 26.73 km/h

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The is organic compound with the correct chemical formula C4H9O2.

A model is a representation of reality. A model serves the purpose of prediction as well as explanation.

Looking at the model of the molecule we can see that it is the organic compound with the correct chemical formula C4H9O2. The molecule is shown in the image attached to this answer.

Missing parts:

There are several ways to model a compound. One type of model is shown.

What is the chemical formula for the molecule represented by the model?

CHO

C4H9O2

C4H8O

C3H8O2

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The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.

Three equation of motion are:-

Where v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.

In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.

Applying equation 1 to find the initial speed of plane

v = u + at

0 = u + (-6.34 × 5400)   {v=0 as plane will stop after 5400 sec}

u =  6.34 × 5400

u = 34236 m/s

Initial velocity of plane is 34236 m/s

Applying equation 2 to find the displacement of plane in that time period

s = ut + (1/2)at²

s = ( 34236 × 5400 )  - ( (1/2) × 6.34 × 5400² )

s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )

s = 5400 × ( 34236 - 17118 )

s = 5400 × 17118 metres

s = 5.4 × 17118 Km

s = 92437.2 Km

Distance travelled by plane is 92437.2 Km

So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.

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R=√20²+40²+2.20.40 cos 50

R=55 N

R/sin β = F/sin α

55/sin 50 = 40/sin α

α = 33°

Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.

Given,

Length, L = 4 m

Outer diameter, D = 300mm, D= 0.3 m

Thickness, t = 50 mm, t = 0.05 m

Stress produced, σ = 75000 kN/m²

Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²

Calculating the diameter of the cylinder,

Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)

d= 0.2 m

(i) Magnitude of the load P:

Using the relation, σ =P/A

P = σ × A = 75000 × π /4 (D² – d² )

P= 75000 × π/4 (0.3² – 0.2²)

P= 75000 × π/4 (0.09 – 0.04)

P = 2945.2 kN

Hence, Magnitude of the load P is 2945.2 kN.

(ii) Longitudinal strain produced, e :

Using the relation, Strain, (e) = stress/E

e= 75000/(1.5 x 10⁸)= 0.0005

Hence, the Longitudinal strain produced is 0.0005.

(iii)Total decrease in length, dL:

The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.

Strain = change in length/original length

e= dL/L

0.0005 = dL/4

dL = 0.0005 × 4m = 0.002m=2mm

Hence,the decrease in length is 2 mm.

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The value of x given the data from the question is 0.86 m

From the data given above, we can s determine the energy as follow:

E = mgh

E = 3 × 9.8 × 5

E = 147 J

E = ½Ke²

147 = ½ × 400 × x²

147 = 200 × x²

Divide both side by 200

x² = 147 / 200

Take the square root of both sides

x = √( 147 / 200)

x = 0.86 m

Since the block came to rest, there is no motion

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