A clarinetist, setting out for a performance, grabs his 3.450 kg3.450 kg clarinet case (including the clarinet) from the top of the piano and carries it through the air with an upward force of 38.92 N.38.92 N. Find the case's vertical acceleration. Indicate an upward acceleration as positive and a downward one as negative.

Answer:

d = 1.02 m

Explanation:

By energy conservation we can find the speed by which ball will leave the ramp

[tex]U_i + KE_i = U_f + KE_f[/tex]

here we know that

[tex]mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2[/tex]

here we have

[tex]h_1 = 1.17 m[/tex]

[tex]h_2 = 0.298 m[/tex]

so we have

[tex](9.8)(1.17) = (9.8)(0.298) + \frac{1}{2} v^2[/tex]

[tex]v = 4.13 m/s[/tex]

now the time taken by the block to reach the ground is given by

[tex]h_2 = \frac{1}{2}gt^2[/tex]

[tex]t = \sqrt{\frac{2h_2}{g}}[/tex]

[tex]t = \sqrt{\frac{2(0.298)}{9.8}}[/tex]

[tex]t = 0.25 s[/tex]

now the distance covered by it is given as

[tex]d = 0.25 \times 4.13[/tex]

[tex]d = 1.02 m[/tex]

Answer:

The elastic deformation is 0.00131.

Explanation:

Given that,

Force F = 44500 N

Cross section [tex]A =15.2mm\times19.1\ mm[/tex]

We Calculate the stress

Using formula of stress

[tex]\sigma=\dfrac{F}{A}[/tex]

Where, F = force

A = area of cross section

Put the value into the formula

[tex]\sigma=\dfrac{44500}{15.2\times10^{-3}\times19.1\times10^{-3}}[/tex]

[tex]\sigma=153.27\times10^{6}\ N/m^2[/tex]

We need to calculate the strain

Using formula of strain

[tex]Y=\dfrac{\sigma}{\epsilon}[/tex]

[tex]epsilon=\dfrac{\sigma}{Y}[/tex]

Where,

[tex]\sigma[/tex]=stress

Y = young modulus of copper

Put the value into the formula

[tex]\epsilon=\dfrac{153.27\times10^{6}}{117\times10^{9}}[/tex]

[tex]\epsilon =0.00131[/tex]

Hence, The elastic deformation is 0.00131.

Answer:

Net force, F = 7 N

Explanation:

It is given that,

Initial kinetic energy of the car, [tex]K_i=3\ J[/tex]

Final kinetic energy of the car, [tex]K_f=10\ J[/tex]

Distance, d = 1 m

We need to find the average net force on the car during this interval. It is given by using the work energy theorem as :

[tex]W=\Delta K[/tex]

[tex]W=K_f-K_i[/tex]

Also, W = F.d    d = distance and F = net force

[tex]F.d=K_f-K_i[/tex]

[tex]F=\dfrac{K_f-K_i}{d}[/tex]

[tex]F=\dfrac{10\ J-3\ J}{1\ m}[/tex]

F = 7 Newton

So, the average net force on the car during this interval is 7 newton. Hence, this is the required solution.

The average net force on a radio-controlled car that increased its kinetic energy from 3 J to 10 J over a distance of 1 m is calculated using the work-energy theorem and found to be 7 newtons.

The student's question asks for the calculation of the average net force on a radio-controlled car that increased its kinetic energy from 3 J to 10 J over a distance of 1 m. To find this, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The work done on the car, which equals the change in kinetic energy, is:

Since work is also defined as the force times the distance (W = F * d) and we know the work done (7 J) and the distance (1 m), we can solve for the force (F):

Therefore, the average net force on the car during this interval was 7 newtons.

Answer:

[tex]E_{net} = 3.6 \times 10^4 N/C[/tex]

Explanation:

As the two charges Q1 and Q2 are placed at some distance apart

so the electric field at mid point will be twice the electric field due to one charge

Because here the two charges are of opposite sign so here the electric field at mid point will be added due to both

so here we have

[tex]E_{net} = 2E[/tex]

[tex]E_{net} = 2(\frac{kQ}{r^2})[/tex]

distance of mid point from one charge is given as

[tex]r = \frac{\sqrt{1^2 + 1^2}}{2}[/tex]

[tex]E_{net} = 2 (\frac{(9\times 10^9)(1\times 10^{-6})}{(\frac{1}{\sqrt2})^2}[/tex]

[tex]E_{net} = 3.6 \times 10^4 N/C[/tex]

Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = [tex]\frac{q_{1}}{t}[/tex] = [tex]\frac{0.004288}{1}[/tex] = 0.004288 A

i₂ = Current due to Cl⁻ ions = [tex]\frac{q_{2}}{t}[/tex] = [tex]\frac{0.006272}{1}[/tex] = 0.006272 A

Current passing between the electrodes is given as

i = i₁ + i₂

i = 0.004288 + 0.006272

i = 0.01056 A

i = 10.6 x 10⁻³ A

i = 10.6 mA

A current of 10.56 mA passes through a sodium chloride solution causing  2.68 × 10¹⁶ Na⁺ ions and 3.92 × 10¹⁶ Cl⁻ ions to arrive at their respective electrodes in 1.00 s.

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space.

2.68 × 10¹⁶ Na⁺ ions (1.60 × 10⁻¹⁹ C/ion) arrive at the negative electrode in 1.00 s.

I₁ = 2.68 × 10¹⁶ ion × (1.60 × 10⁻¹⁹ C/ion)/ 1.00 s × (10³ mA/1 A) = 4.29 mA

3.92 × 10¹⁶ Cl⁻ ions (1.60 × 10⁻¹⁹ C/ion) arrive at the positive electrode in 1.00 s.

I₂ = 3.92 × 10¹⁶ ion × (1.60 × 10⁻¹⁹ C/ion)/ 1.00 s × (10³ mA/1 A) = 6.27 mA

I = I₁ + I₂ = 4.29 mA + 6.27 mA = 10.56 mA

A current of 10.56 mA passes through a sodium chloride solution causing  2.68 × 10¹⁶ Na⁺ ions and 3.92 × 10¹⁶ Cl⁻ ions to arrive at their respective electrodes in 1.00 s.

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Answer:

Explanation:

To find the force between the two charges, we required the charges on each object and the distance between the two objects.

In the question, the distance between two charges is not given so we cannot find the charge on each sphere.

If the force is attractive, then the charges are opposite in nature and if the force is repulsive then the charges are same in nature.

Doubling the plate separation of a parallel-plate capacitor would double the potential difference to maintain the same electric field strength, as the electric field in a capacitor is proportional to the charge on the plates. Therefore, the answer is E. none of the above.

If the plate separation of an isolated charged parallel-plate capacitor is doubled, the correct effect on the capacitor's characteristics from the options provided is: the potential difference is doubled. This is because the electric field (E) in a parallel-plate capacitor is given by E = V/d, where V is the potential difference and d is the separation between the plates. When the plate separation is doubled, the electric field remains unchanged (since the charge remains the same and the electric field strength is directly proportional to the charge on the plates). As a result, the potential difference must also double to maintain the same electric field strength.

The charge on each plate does not change, and therefore, neither does the surface charge density, since it is defined as the charge per unit area (Q/A), and there is no indication that the area changes. Therefore, the correct answer is E. none of the above

Answer:

[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]

Explanation:

For pendulum A: Length = L and gravity = g

The frequency of pendulum A is given by

[tex]f = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex]

Here, f is the frequency, L be the length

[tex]f_{A} = \frac{1}{2\pi }\sqrt{\frac{g}{L}}[/tex]     ... (1)

For pendulum B: Length = 2L, gravity = g

The frequency of pendulum B is given by

[tex]f_{B} = \frac{1}{2\pi }\sqrt{\frac{g}{2L}}[/tex]   .... (2)

Divide equation (1) by (2)

[tex]f_{B}: f_{A} = \sqrt{\frac{2}{1}}[/tex]

Answer:

9

Explanation:

As the net force on Q is zero, so the force on Q due to q1 is balanced by the force on Q due to q2.

Let the force on Q due to q1 is F1 and force on Q due to q2 is F2.

F1 = F2

K Q q1 / (0.5 a)^2 = K Q q2 / (1.5 a)^2

q1 / 0.25 = q2 / 2.25

q2 / q1 = 2.25 / 0.25

q2 / q1 = 9

Answer:

34.3 N and 15.7 N

Explanation:

[tex]F_{left}[/tex]  = force on the left end of the rod

[tex]F_{right} [/tex]  = force on the right end of the rod

M = mass of Halloween decoration = 3 kg

[tex]F_{h}[/tex]  = weight of the Halloween decoration = Mg = 3 x 9.8 = 29.4 N

m = mass of rod = 2.1 kg

[tex]F_{r}[/tex]  = weight of the rod = mg = 2.1 x 9.8 = 20.6 N

From the force diagram, using equilibrium of torque about A

[tex]F_{h}[/tex] (AB) + [tex]F_{r}[/tex] (AC) = [tex]F_{right}[/tex] (AD)

(29.4) (21) + (20.6) (57) = [tex]F_{right}[/tex] (114)

[tex]F_{right}[/tex]  = 15.7 N

Using equilibrium of force along the vertical direction

[tex]F_{right}[/tex] + [tex]F_{leftt}[/tex] =  [tex]F_{h}[/tex] + [tex]F_{r}[/tex]

15.7 +  [tex]F_{leftt}[/tex] = 29.4 + 20.6

[tex]F_{leftt}[/tex] = 34.3 N

To find the force a curtain rod support must hold, calculate the torques due to the rod's weight and the decoration's weight, considering their distances from the support. Then, balance these torques with the force at the other support. The calculated force for one support is approximately 15.7 newtons.

To calculate the force one of the curtain rod supports must be able to hold, we must consider the torques about one end of the rod and include the mass of the rod and the Halloween decoration. The rod and the decoration together exert a torque about the pivot point due to gravity. The support must exert an equal and opposite torque to keep the system in equilibrium. We do this by summing the torques and setting them to zero, since the system is not rotating.

The total torque ( au) about the right support due to the rod and the decoration can be calculated using  au = r  imes F where r is the distance from the pivot and F is the force due to gravity (weight). The weight of the rod acts at its center of mass, which is at the midpoint of the rod, while the weight of the decoration acts at 21 cm from the right end. So, the distances from the pivot (right end) to the centers of mass are:

Rod: 57 cm (half of the total length of 114 cm)

Decoration: 21 cm

The weights are:

Rod: 2.1 kg  imes 9.8 m/s2 = 20.58 N

Decoration: 3 kg  imes 9.8 m/s2 = 29.4 N

The torques about the right support are:

Rod: 57 cm  imes 20.58 N = 1173.06 N ext{cm}

Decoration: 21 cm  imes 29.4 N = 617.4 N ext{cm}

Since the rod is in equilibrium, the torques must balance, meaning the support at the left end must provide an upwards force resulting in a torque that equals the sum of the other two torques. Let F be the force at the left end support. It must satisfy:

F  imes 114 cm = 1173.06 N ext{cm} + 617.4 N ext{cm}

F = (1173.06 + 617.4) / 114

F ≈ 15.7 N

Therefore, the force one of the curtain rod supports needs to be able to hold is approximately 15.7 newtons.

Answer:

Is required 5.26 min to raise the temperature.

Explanation:

R= 18 ohms

V= 120 volts

m= 0.71 kg

C= 4186 J/kg °C

T1= 15°C

T2= 100 °C

Q= m * C * (T2-T1)

Q= 252.62 *10³ J

V/R= I

I= 6.66 A

P= I² * R

P= 800 W = 800 J/s

P= Q/t

t= Q/P

t= 315.77 s = 5.26 min

Answer:

At r= 0.76 m measured from the axis of rotation, does the tangential acceleration equals the acceleration of the gravity.

Explanation:

at=g= 9.8 m/s²

α= 12.8 rad/s²

r= at/α

r= 0.76m

Explanation:

N. B separator is the correct ande

Answer:

[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

[tex]F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}[/tex] (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

[tex]F=m\frac{v^2}{r}[/tex] (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

[tex]k\frac{Ze^2}{r^2}=m\frac{v^2}{r}[/tex]

And solving for v, we find an expression for the speed of the electron:

[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]

The speed of an electron in a circular orbit around a nucleus is determined by equating the Coulomb force and the centripetal force, resulting in the expression v = sqrt(kZe^2/mr).

The speed v of an electron in a circular orbit around a nucleus can be found by equating the electrostatic force to the centripetal force required for circular motion. The electrostatic force, due to the Coulomb's interaction, between the electron and the nucleus is given by F = k(Ze)(-e)/r^2, where k is Coulomb's constant, Z is the atomic number, e is the magnitude of the charge of an electron, and r is the radius of the orbit. On the other hand, the centripetal force needed to keep the electron in circular motion is F = mv^2/r where m is the mass of the electron and v is its speed.

Setting the two expressions equal gives the equation for the electron's speed v:
k(Ze)(-e)/r^2 = mv^2/r
Solving for v results in the expression:
v = sqrt(kZe^2/mr)

This equation shows that the electron's speed in its orbit depends on the atomic number Z, Coulomb's constant k, the electron's mass m, and the orbit radius r.

Answer:

8.64 m/s

Explanation:

v1 = 8 m/s, v2 = ?, P2 - P1 = 5338 Pa, density of water, d = 1000 kg/m^3

By the use of Bernoulli's theorem

P 1 + 1/2 x d x v1^2 = P2 + 1/2 x d x v2^2

P2 - P1 = 1/2 x d x (v2^2 - v1^2)

5338 = 0.5 x 1000 x (v2^2 - 64)

10.676 = v2^2 - 64

v2^2 = 74.676

v2 = 8.64 m/s  

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

Answer : The balanced two-half reactions will be,

Oxidation half reaction (anode) : [tex]Mg(s)\rightarrow Mg^{2+}(aq)+2e^-[/tex]

Reduction half reaction (cathode) : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]

Thus the overall reaction will be,

[tex]Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)[/tex]

Explanation :

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

The given redox reaction occurs between the magnesium and cadmium.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The balanced two-half reactions will be,

Oxidation half reaction (anode) : [tex]Mg(s)\rightarrow Mg^{2+}(aq)+2e^-[/tex]

Reduction half reaction (cathode) : [tex]Cd^{2+}(aq)+2e^-\rightarrow Cd(s)[/tex]

Thus the overall reaction will be,

[tex]Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)[/tex]

half-reactions

cathode : Cd²⁺ (aq) + 2e⁻ ---> Cd (s)

anode :  Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)

Cell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.

Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)

[tex] \large {\boxed {\bold {E ^ osel = E ^ ocatode -E ^ oanode}}} [/tex]

or:

E ° cell = E ° reduction-E ° oxidation

(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)

The value of E cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode

In reaction:

Cd²⁺ + Mg → Cd + Mg²⁺

half-reactions

a balanced cell reaction

Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)

The standard cell potential

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The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

I = 800.00A

Δt = 1.2min =  72s

Plug in the values and solve for Q:

800.00 = Q/72

Q = 57600C

Answer:

72.75 kg

Explanation:

[tex]F[/tex] = force applied on a box = 750 N

[tex]m[/tex] = mass of the box

[tex]N[/tex] = Normal force on the box

[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.66

From the force diagram, force equation along the vertical direction is given as

[tex]N = F Sin25 + mg[/tex]

[tex]N = 750 Sin25 + mg[/tex]                                                         eq-1

Static frictional force is given as

[tex]f_{s} = \mu _{s} N[/tex]

using eq-1

[tex]f_{s} = \mu _{s} (750 Sin25 + mg)[/tex]

For the box to move,

[tex]F Cos25 = f_{s}[/tex]

[tex]750 Cos25 = \mu _{s} (750 Sin25 + mg)[/tex]

[tex]750 Cos25 = (0.66) (750 Sin25 + m (9.8))[/tex]

m = 72.75 kg

Answer:

1.45255 x 10⁸ m/s

Explanation:

q  = magnitude of charge on the electron = 1.6 x 10⁻¹⁹ C

m = mass of the electron = 9.1 x 10⁻³¹ kg

v = speed of the electron

ΔV = potential difference = 60,000 Volts

Using conservation of energy

Kinetic energy gained by the electron = Electric potential energy

(0.5) m v² = q ΔV

(0.5) (9.1 x 10⁻³¹ ) v² = (1.6 x 10⁻¹⁹) (60,000)

v = 1.45255 x 10⁸ m/s

Answer:

11.54 ft - lb

Explanation:

F = 18 lb, x = 8 in = 8 / 12 = 0.66 ft

F = k x

k = F / x = 18 / 0.66 = 27.27 lb/ft

y = 11 in = 11 / 12 ft = 0.92 ft

Work done = 1 /2 x k x  y^2

W = 0.5 x 27.27 x 0.92 x 0.92 = 11.54 ft - lb

Work done of spring is the product of average force and the displacement.

The work required in stretching spring from its natural length to 11 in. beyond its natural length is 11.54 ft-Ib.

Work done of spring is the product of average force and the displacement. It can be given as,

[tex]W=\dfrac{1}{2}kx^2[/tex]

Here, [tex]k[/tex] is the spring constant. The spring constant can be given as,

[tex]k=\dfrac{F}{x}[/tex]

Here, [tex]F[/tex] is the force and [tex]x[/tex] is the displacement of spring.

Given information-

The value of the force is 18 Ib.

The length of spring stretched is 8 in beyond its natural length.

Change the length in feet as,

[tex]x=\dfrac{8}{12} \\x=0.66\rm ft[/tex]

Put the value in the above formula as,

[tex]k=\dfrac{18}{0.66}\\k=27.27 \rm Ib/ft[/tex]

Work W is required in stretching it from its natural length to 11 in. beyond its natural length, Change this length in feet as,

[tex]y=\dfrac{11}{12} \\y=0.94\rm ft[/tex]

Put the values in the formula of work done as,

[tex]W=\dfrac{1}{2}\times27.27\times (0.92)^2\\W=11.54 \rm ft-Ib[/tex]

Thus the work required in stretching spring from its natural length to 11 in. beyond its natural length is 11.54 ft-Ib.

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Answer:

B) 2.18 m/s²

Explanation:

M = Original mass of earth

R = Original radius of earth

g = original acceleration due to gravity of earth = 9.8 m/s²

M' = New mass of earth = 2 M

R' = New radius of earth = 3 R

g = original acceleration due to gravity of earth = 9.8 m/s²

Original acceleration due to gravity of earth is given as

[tex]g=\frac{GM}{R^{2}}[/tex]

[tex]9.8 =\frac{GM}{R^{2}}[/tex]                               eq-1

g' = new acceleration due to gravity of earth

New acceleration due to gravity of earth is given as

[tex]g' =\frac{GM'}{R'^{2}}[/tex]

[tex]g' =\frac{G(2M)}{(3R)^{2}}[/tex]

[tex]g'=\left ( \frac{2}{9} \right )\frac{GM}{R^{2}}[/tex]

Using eq-1

[tex]g'=\left ( \frac{2}{9} \right )(9.8)[/tex]

g' = 2.18 m/s²

Final answer:

When the Earth's radius is tripled and its mass is doubled, the new surface gravitational acceleration would be calculated using the gravitational acceleration formula and would result in a new acceleration of B) 2.18 m/s².

Explanation:

To find the new surface gravitational acceleration (g') when the radius of the Earth is tripled and its mass is doubled, we use the formula for gravitational acceleration:

g' = (G × new mass) / (new radius)²

Where:

Substituting the new mass and radius into the formula, we get:

g' = (G × 2M) / (3R)²

g' = (2G × M) / (9R²)

g' = (2/9) × (G × M / R²)

Since G × M / R² is the original gravitational acceleration (g) of Earth, which is 9.8 m/s², we can now substitute:

g' = (2/9) × 9.8 m/s²

g' = 2.18 m/s²

So the new surface gravitational acceleration would be 2.18 m/s², which matches option B.

Answer:

[tex]l =1mm[/tex]

Explanation:

Given:

Rate of heat transfer, P = 150 W

Body surface Area, A = 1.5 m²

Temperature difference, ΔT = 0.50°C

Also,

The rate of heat transfer, P is given as:

[tex]P = \frac{kA\Delta T}{l}[/tex]

Where,

l =length of material (or here it isaverage distance of the capillaries below the skin surface)

k = Thermal conductivity

Here the transfer of heat is through the skin. Thus, k for human tissue is given as 0.2

substituting the values in the above equation, we get

[tex]150 = \frac{0.2\times 1.5\times 0.50}{l}[/tex]

or

[tex]l = \frac{0.2\times 1.5\times 0.50}{150}[/tex]

or

[tex]l = 1\times 10^{-3}m=1mm[/tex]

Using the formula for thermal conduction, it is estimated that the capillaries lie about 3 mm below the skin surface. This is a simplification, actual distances can vary based on specific factors.

To estimate the average distance between the capillaries and skin surface, we can use the formula for thermal conduction, which states that heat flow equals the thermal conductivity constant (k) times the surface area of the skin (A) times the temperature difference (ΔT), divided by the thickness of the skin (d), or Q = k*A*ΔT/d.

Assuming that human skin has a thermal conductivity similar to water (k~0.6 W/mK), we can rearrange the formula to solve for d: d = k*A*ΔT/Q. Plugging in the given values, we get d = (0.6 W/mK * 1.5 m^2 * 0.50 ºC) / 150 W, which simplifies to approximately 0.003 m, or 3 mm.

This estimate suggests that, on average, the capillaries lie about 3 mm below the skin surface. However, this is a simplification and actual distances can vary based on factors such as the specific region of the body, individual physiology, and more. Thermal conduction is just one mechanism of heat transfer in the body, along with convection and radiation.

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Answer:

The height of the bag is 1.466 m.

Explanation:

Given that,

Acceleration of gravity [tex]g=9.8\ m/s^2[/tex]

Pressure = 14800 Pa

Specific gravity = 1.03

We need to calculate the density

Using formula of specific gravity

[tex]\rho_{s}=\dfrac{\rho}{\rho_{w}}[/tex]

[tex]rho=\rho_{s}\times{\rho_{w}}[/tex]

Where, [tex]\rho[/tex] = density of solution

[tex]\rho_{w}[/tex] = density of water

Put the value in to the formula

[tex]\rho=1.03\times1000[/tex]

[tex]\rho=1030\ kg/m^3[/tex]

We need to calculate the height

Using formula of pressure

[tex]P=\rho gh[/tex]

[tex]h=\dfrac{P}{\rho g}[/tex]

Where, P = pressure

g = acceleration due to gravity

h = height

Put the value into the formula

[tex]h = \dfrac{14800}{1030\times9.8}[/tex]

[tex]h=1.466\ m[/tex]

Hence, The height of the bag is 1.466 m.

The minimum height of the bag of glucose solution must be 1.46 meters above the entry point into the vein for the fluid to just enter, calculated using the pressure equation P = hρg.

In order to infuse the glucose into a vein, the pressure must be greater than the pressure in the vein. This can be achieved by finding the height of the fluid that corresponds to this greater pressure. Using the pressure equation P = hρg, where P is the pressure, h is the height of the fluid, ρ is the density of the fluid, and g is the acceleration due to gravity. In the given question we can solve for the height h:

h = P/(ρg)

Let's substitute the given values into the equation. The density ρ of the glucose solution is 1.03 times water's density since its specific gravity is given as 1.03. The density of water is 1000 kg/m³, so the density of the glucose solution is 1030 kg/m³. So, we get:

h = 14800 Pa / (1030 kg/m³ * 9.8 m/s²)

This calculates the minimum height of the collapsible plastic bag to be 1.46 meters (rounded to two decimal places) above the entry point into the vein for the fluid to just enter the vein.

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Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

[tex]v_2=-\dfrac{m_1v_1}{m_2}[/tex]

[tex]v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}[/tex]

[tex]v_2=-1.47\ m/s[/tex]

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

Answer:

The final velocity of the ball, v = 9.9 m/s

Explanation:

It is given that,

A ball is dropped 5 meters from rest, h = 5 meters

We need to find the final velocity of the ball. It can be calculated using the conservation of energy as :

[tex]KE_i+KE_f=PE_i+PE_f[/tex]

Initial kinetic energy, [tex]KE_i=0\ (rest)[/tex]

Final kinetic energy, [tex]KE_f=\dfrac{1}{2}mv^2[/tex]

Initial potential energy, [tex]PE_i=mgh[/tex]

Final potential energy, [tex]PE_f=0[/tex] (at ground, h = 0)

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\ m/s^2\times 5\ m}[/tex]

v = 9.89 m/s

or

v = 9.9 m/s

So, the final velocity of the ball is 9.9 m/s. Hence, this is the required solution.

Answer:

Horizontal component of velocity shall be [tex]v_{fx}=v_{o}cos(70^{o})[/tex]

Explanation:

Since  the given projectile motion is under the influence of gravity alone which acts in vertical direction only and hence the acceleration shall act in vertical direction only and correspondingly if air resistance is neglected the acceleration in the horizontal direction shall be zero.

For zero acceleration in the horizontal direction the velocity in horizontal direction shall not change.

Mathematically

[tex]v_{ix}=v_{fx}[/tex]

We have initial horizontal velocity =[tex]v_{ix}=v_{o}cos(70^{o})[/tex]

Thus this shall remain constant throughout the course of the motion.

Answer:

120 rpm

Explanation:

I1 = 105 kgm^2, I2 = 70 kgm^2

f1 = 80 rpm, f2 = ?

Let the angular speed be f2 when his arms are tucked.

If no external torque is applied, then the angular momentum remains constant.

L1 = L2

I1 w1 = I2 w2

I1 x 2 x 3.14 x f1 = I2 x 2 x 3.14 x f2

105 x 80 = 70 x f2

f2 = 120 rpm

When a battery is connected in a series to a resistor, its power quadruples. So, the power that a resistor dissipating 0.5 W originally would dissipate when connected to two identical batteries in series would be 2.00 W.

In the world of physics, power dissipated by a resistor in a circuit is governed by the equation P = I2R, where P is the power, I is the current, and R is the resistance. If the resistance stays unchanged and you double the voltage (by adding identical battery in series), the current through the circuit doubles as well. Thus, with the power quadrupling as a result of two times the current squared (since 22 = 4), the resistor would now dissipate 2.00 W of power.

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When two identical batteries are connected in series with a resistor originally dissipating 0.50 W, the power dissipation increases to 2.0 W. This is because the total voltage supplied to the circuit is doubled, leading to a quadratic increase in power dissipation.

To find the new power dissipation, we start by applying Ohm's law and the power formula. Initially, the power dissipated is given by P = IV. Since power can also be expressed as P = V²/R and the resistor dissipates 0.50 W:

P = V²/R = 0.50 W

Now, connecting two identical batteries in series doubles the voltage:

[tex]V_{new[/tex] = 2V

The new power dissipation is then:

[tex]P_{new[/tex] = ([tex]V_{new[/tex])² / R

[tex]P_{new[/tex] = (2V)² / R

[tex]P_{new[/tex] = 4V² / R

[tex]P_{new[/tex] = 4 * 0.50 W

[tex]P_{new[/tex] = 2.0 W

Therefore, the resistor now dissipates 2.0 W of power.

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

[tex]\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k[/tex]

Part b)

[tex]\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1[/tex]

[tex]\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}[/tex]