A 550-g squirrel with a surface area of 945 cm2 falls from a 4.0-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
m/s

What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
m/s

The net current in the conductors and the value of the line integral

This is further explained below.

Generally,

To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).

[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}[/tex]

[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]

B)

In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same at 3.2 *10^4 Tm

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The mass of the planet Mar, given the data from the question is 6.45×10²³ Kg

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by G

M = gr² / G

M = (3.724 × 3400000²) / 6.67×10¯¹¹

M = 6.45×10²³ Kg

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Answer:

Higher levels of organization are built from lower levels.

Molecules combine to form cells.

Cells combine to form tissues.

Tissues combine to form organs.

Organs combine to form organ systems,

and organ systems combine to form organisms.

Explanation:

Hope it helps.

0.02020 ohm is the resistance of a carbon rod at 25.8 ∘C if its resistance is 0.0200 Ω at 0.0 ∘C.

A resistor is an electrical component that controls or restricts how much electrical current can pass across a circuit in an electronic device. A specified voltage can be supplied via resistors to an active device like a transistor.

The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:

R = R0[1+ alpha(T-T0)]  where:

R is the new resistance we are looking for

alpha is the temperature coefficient of resistance. For carbon rod, alpha = ₋ 4.8 x [tex]10^{-4}[/tex](1/°c)

T0 is the standard temperature =25.8°C

R0 is the resistance at T0 = 0.0200 ohms

T is the temperature at which we want to get R = 0

Substitute in the equation to get R as follows:

R = 0.0200 [1+( ₋ 4.8 x [tex]10^{-4}[/tex]) (0-25.8)] = 0.02020 ohm

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(a) The acceleration of the 1,100-kg boat is 0.341 m/s².

(b) The distance covered by the boat is 68.2 m.

(c) The speed of the boat is 6.82 m/s.

Net force on the boat = 1,575 N - 1,200 N = 375 N

F(net) = ma

a = F(net)/m

a = 375/1100

a = 0.341 m/s²

s = ut + ¹/₂at²

s = 0 + ¹/₂(0.341)(20)²

s = 68.2 m

v = u + at

v = 0 + 0.341(20)

v = 6.82 m/s

Thus, the acceleration of the 1,100-kg boat is 0.341 m/s², the distance covered by the boat is 68.2 m and the speed of the boat is 6.82 m/s.

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Ant is performing a work

what is work?

Work is the force applied on an individual with respect to displacement.

Work = Force × displacement

Unit is Nm

Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.

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The ant works, but the elephant does not.

Work done = Force × Displacement.

If there are ants and houseflies,

Ants drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.

In the case of the elephant and the tree,

When the elephant pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.

Work done = Force × Displacement

= Force × 0

= 0

Therefore,

The ant works, but the elephant does not.

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The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

                                     [tex]a=0\\\alpha =35^0\\T=75N\\m=57kg[/tex]

Here, acceleration will be equal to zero, because the velocity is given as constant.

                                      [tex]ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha[/tex]

                   [tex]ma= Tcos\alpha -k(mg-Tsin\alpha )[/tex]

                    [tex]0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119[/tex]

Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.

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The sled's coefficient of friction with the snow is 0.119.

We must understand the friction in order to choose the solution.

                               [tex]\alpha =35\\T=75N\\m=57kg\\a=0[/tex]

Because the velocity is specified as constant in this case, the acceleration will be equal to zero.

                             [tex]ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha[/tex]

                       [tex]ma=Tcos\alpha -k(mg-Tsin\alpha )[/tex]

                                   [tex]k=0.119[/tex]

As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.

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C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.

Center of gravity is the point from which the weight of a body or system may be considered to act.

Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.

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Checking the oil will reduce the possibility of having to rebuild or replaced the engine.

what is an engine ?

A device created to transform one or more sources of energy into mechanical energy is known as an engine or motor. Potential energy, heat energy, chemical energy, electric potential, and nuclear energy are all forms of energy that are readily available.

The lubricating function of engine oil is crucial. It shields and stops all the moving parts from rubbing against one another. Metal-on-metal wear would quickly kill your engine without lubrication.

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SHM can be acquired by perpendicular projection of uniform circular motion of a particle on its diameter such a particle is called reference particle and its circular path is called reference circle.

A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the term for conservation of energy to determine the maximum height of the block.

As the pendulum swings downward, gravity converts this potential energy into kinetic energy, so that at the bottom of the swing, the pendulum bob has zero potential energy, and its kinetic power, (1/2)mv2, equals the initial potential energy (mgh). (So the velocity, v, equals √(2gh).)

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The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².

Acceleration is the change in velocity with time.

The maximum acceleration is obtained by taking moments about the tipping point of rotation.

F₂ * 1.58 m = F₁ * 0.67 m

where

F₂ is tipping force = mass * acceleration, a

F₁ is weight = mass * acceleration due to gravity, g

The weight acts at a distance half the width of the refrigerator = 30 cm or 0.3 m

Height of refrigerator is 158 cm 0r 1.58 cm

m * a * 1.58 = m * 9.81 * 0.30

a = 1.86 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².

In conclusion, if the maximum acceleration is exceeded, the refrigerator will tip over.

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The weight of the body is obtained as 53.9 N.

The term weight refers to the product of the mass and the acceleration due to gravity.

Now we have the mass  of the body as 5.5kg and the acceleration due to gravity as 9.8 m/s^2.

It the follows that the weight is;

W = mg = 5.5kg *  9.8 m/s^2 = 53.9 N

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Answer:

The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.

[tex]s\frac{d}{t}[/tex]

Explanation:

hope this helps you my friend

Answer:  Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure. The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). Then, at an angle 135.48°, the planet Y rotated through during this time.

Explanation: To find the answer, we need to know about the Kepler's third law of planetary motion.

                                      T² ∝ r³

                 [tex]\frac{r_1}{r_2}=\frac{4}{5} \\\frac{T_1}{T_2} =(\frac{r_1}{r_2})^{3/2}\\[/tex]

                  [tex]T=\frac{2\pi }{w}[/tex] where, w is the angular velocity.

                           [tex]\alpha[/tex] =wt.

                   [tex]\frac{T_x}{T_y}=\frac{w_y}{w_x}\\thus,\\\frac{w_y}{w_x}=(\frac{r_1}{r_2})^{3/2}[/tex]

                     [tex]\frac{\alpha _y}{\alpha _x} = (\frac{r_1}{r_2})^{3/2}\\where,\\\alpha _x=\frac{88^0}{5 yrs} because,\\here, angle \beta_x =88^0.\\[/tex]

                     [tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} )^{3/2}=\frac{135.48^0}{5yrs} .\\thus,\\\beta _y=135.48^0.[/tex]

Thus, we can conclude that the angle rotated by planet Y during 5 yrs will be 135.48 degrees.

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The planet Y then rotated through at this time at an angle of 135.48°.

In order to understand the solution, we must be familiar with Kepler's third law of planetary motion.

                              T² ∝ r³

                                  [tex]\frac{R_1}{R_2}=\frac{4}{5} \\\frac{T_1}{T_2}=(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]

                                [tex]T=\frac{2\pi }{w}[/tex]

where w is the angle of rotation per time.

                                     [tex]\alpha[/tex]=wt.

                                   [tex]\frac{w_y}{w_x} =(\frac{4}{5} ) ^{\frac{3}{2} }[/tex]

                                   [tex]\frac{\alpha _y}{\alpha _x}=(\frac{4}{3} ) ^{\frac{3}{2} } , where\\\alpha _x=\frac{88}{5yrs} \\[/tex]

                           [tex]\alpha _y=\frac{88}{5yrs} *(\frac{4}{3} ) ^{\frac{3}{2} }=\frac{135.48}{5yrs}[/tex]

Thus, we may infer that planet Y will revolve at an angle of 135.48 degrees during the course of five years.

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The tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (40 x 9.8)/(2 x sin31)

T = 380.55 N

T + F = W

F = W - T

F = (9.8 x 40) - 380.55

F = 11.45 N

Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

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Answer:

See below

Explanation:

With switches open, the circuit is a simple series circuit ....the ammeters will have the same readings

V = IR

I = V/R = 5 / (10+5+5) = .25 A

b) With S1 closed   5 ohm and 10 ohm in parallel become = 5 *10 / (5+10) = 3.33 ohm

 then the series circuit current becomes  

     5 v / ( 10 + 3.33 + 5 ) = ammeter 1 = .273 amps

            ammeter 2 will get a portion of this ...the smaller resistor will get 2/3 ...the 10 ohm resistor will get 1/3        .273 *   10 / 15 =.182 amps

Both forming images when placed in their respective places best describes  Becquerel’s experiments.

This was conducted by  Henri Becquerel in which he sought to know how uranium salts are affected by light.

He discovered that the salts emits a penetrating radiation and formed an image in the presence of light but didn't form any in darkness.

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The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.

The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.

λ=h/p

Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.

Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.

Therefore, λ=h/(mv)

λ=(6.63×10⁻³⁴)/(0.56×26)

λ=4.55×10⁻³⁵ m.

The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.

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The angle between the two beams inside the glass is mathematically given as

<(BR)= 0.563°

Generally, the equation for an angle for blue is  mathematically given as

[tex]< B= arcsin( sin\theta / i )[/tex]

Therefore

<B= arcsin( sin30 / 1.660 )

<B= 17.53°

For angle for red

<R= arcsin(sin30 / 1.610)

<R= 18.09°

For the angle in between

<(BR)= 18.093 - 17.530

<(BR)= 0.563°

In conclusion,  the angle between the two beams inside the glass

<(BR)= 0.563°

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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

[tex]R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0[/tex]

where [tex]R_1,R_2[/tex] are the magnitudes of the tensions in ropes 1 and 2, respectively;

[tex]R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0[/tex]

where [tex]m=1300\,\rm kg[/tex] and [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].

Eliminating [tex]R_2[/tex], we have

[tex]\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)[/tex]

[tex]R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2[/tex]

[tex]R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2[/tex]

[tex]-R_1 \sin(50^\circ) = -\dfrac{mg}2[/tex]

[tex]R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}[/tex]

Solve for [tex]R_2[/tex].

[tex]\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0[/tex]

[tex]\dfrac{R_2}2 = -mg\cot(110^\circ)[/tex]

[tex]R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}[/tex]

The angle of refraction of the light beam is determined as  16.46 ⁰.

n = sin i / sin r

where;'

Angle between the ray and the normal = incident angle = 24.8⁰

1.48 = sin (24.8) / (sin r)

sin r = sin (24.8) / (1.48)

sin r = 0.283

r = sin ⁻¹(0.283)

r = 16.46 ⁰

Thus, the angle of refraction of the light beam is determined as  16.46 ⁰.

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The angle of the planet is mathematically given as

dY= 704 degrees

With Kepler's third rule, which states that a planet's orbit squared is a function of cubed radius, we can prove that this is the case.

Generally, the equation for the period is  mathematically given as

(periodX / periodY)^2 = (radius X / radius Y)^3

Therefore

(pX / pY)^2 = 4^3

(pX / pY)^2 = 64

\sqrt{(pX / pY )^2}= \sqrt{64}

(pX / pY=8

In conclusion, Because it takes 8 times longer to complete one orbit on planet X, planet Y travels 8 times farther than planet X does in the same time period...

planet Y travels ;

dY=8 * 88.0

dY= 704 degrees

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Answer:

[tex]\alpha=arccos[\frac{(a^2+b^2)(n-1)}{2ab(n+1)}].[/tex]

Explanation:

1) for A+B: a²+b²-2abcos(π-α);

2) for A-B: a²+b²-2abcos(α);

3) according to the condition (A+B):(A-B)=n, then

[tex]n=\frac{a^2+b^2-2abcos(\pi-a)}{a^2+b^2-2abcos(a)}; \ = > \ n=\frac{a^2+b^2+2abcos(a)}{a^2+b^2-2abcos(a)}; \ = > \ cos(\alpha)=\frac{(a^2+b^2)(n-1)}{2ab(n-1)}.[/tex]

(a) The location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

(b) The magnitude and direction of the electric field at the top corner due to the two charges at the base is  1.732 kq/a².

The electric field is zero at the center of the equilateral triangle whose magnitude is equal to √3a/6.

E = E₁ + E₂

where;

E = kq/a²[(cos 60i + sin 60j) + (-cos 60i + sin 60j)]

E = kq/a²[2(sin 60j)] = 1.732 kq/a²

Thus, the location of a point where the electric field is zero is at the center of the triangle which is equal to ¹/₆√3a.

The magnitude and direction of the electric field at the top corner due to the two charges at the base is  1.732 kq/a².

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Answer:

11.31 [m/s].

Explanation:

1. the required velocity can be calculated according to

[tex]V=\frac{V_{horizontal}}{sin45};[/tex]

2. according to the formula above:

V=8*1.41≈11.3137085 [m/s].

The tension, T in the cable is equal to 323.5 N.

Tension is force exerted by a cable or string on another object usually a weight suspended from the cable or string

The tension in the cable is found this:

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

T = 241.68/0.747

T = 323.5 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

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The wave interaction that is shown in the photo is refraction as light moves from air to water.

Refraction refers to the change in the frequency of a wave and the direction of the wave as it moves from one medium to another. We know that waves makes a body under water to look slightly different than when it is in air.

Thus, the wave interaction that is shown in the photo is refraction as light moves from air to water.

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Answer:

a = 7.5

Explanation:

m = f/a

f = 30

m = 4

Thus;

a = f/m

a = 30/4

a = 7.5

Answer:

Explanation: