A man is holding a 6.0-kg (weight = 59 N) dumbbell at arm's length, a distance of 0.56 mfrom his shoulder. What is the torque on the shoulder joint from the weight of the dumbbell if thearm is held at 15° above the horizontal? On the picture, draw the lever arm for this force

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by  [tex]v=\sqrt{\frac{GM}{r}}[/tex], where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

              [tex]v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s[/tex]

  The International Space Station move at 7.22 km/s.      

Answer:

1.1 sec

Explanation:

m = mass of the box = 8 kg

k = spring constant of the spring = 69 N/m

v = initial speed of the box = 1.5 m/s

t = time period of oscillation of box in contact with the spring

Time period is given as

[tex]t = \pi \sqrt{\frac{m}{k}}[/tex]

Inserting the values

[tex]t = (3.14) \sqrt{\frac{8}{69}}[/tex]

t = 1.1 sec

Answer:

1.33 x 10^26 photons

Explanation:

λ = 12 cm = 0.12 m

Δ T = 5 degree C

m = 11 g = 0.011 kg

c = 4 J/g K = 4000 J / kg K

Let the number of photons are n.

Energy of one photon = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

Energy of n photons, E = n h c / λ

This energy is used to raise the temperature of eye, so

n h c / λ = m c Δ T

n = m c Δ T λ / h c

n = (0.011 x 4000 x 5 x 0.12) / (6.63 x 10^-34 x 3 x 10^8)

n = 1.33 x 10^26

Thus, there are 1.33 x 10^26 photons.

To estimate the number of photons that must be absorbed to raise the temperature of an eye by 5.00 °C, calculate the energy required and convert it into the number of photons using Planck's constant and the frequency of the microwaves.

To estimate the number of photons that must be absorbed to raise the temperature of an eye by 5.00 °C, we need to calculate the amount of energy required and then convert it into the number of photons. Here are the steps:

Therefore, approximately 3.30 × 10^11 photons with a wavelength of 12 cm must be absorbed to raise the temperature of an eye by 5.00 °C.

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Answer:

(a) 15.4 second

(b) 769.23 m

Explanation:

u = 0, a = 6.5 m/s^2, v = 100 m/s

(a) Let t be the time taken

Use First equation of motion

v = u + a t

100 = 0 + 6.5 x t

t = 15.4 second

(b) Let height covered is h.

Use third equation of motion

v^2 = u^2 + 2 a h

100^2 = 0 + 2 x 6.5 x h

10000 = 13 x h

h = 769.23 m

Answer:

Charge, [tex]q=2.98\times 10^{-8}\ C[/tex]

Explanation:

It is given that,

Value of electric field, [tex]E=7.5\times 10^5\ V/m[/tex]

Area of parallel plates, [tex]A=45\ cm^2=0.0045\ m^2[/tex]

Distance between two parallel plates, d = 2.45 mm = 0.00245 m

For a parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{\epsilon_oA}{d}[/tex].......(1)

Since, [tex]E=\dfrac{V}{d}[/tex]

V = E . d ............(2)

And [tex]C=\dfrac{q}{V}[/tex].....(3)

From equation (1), (2) and (3) we get :

[tex]\dfrac{q}{V}=\dfrac{\epsilon_oA}{d}[/tex]

[tex]q=\epsilon_o EA[/tex]

[tex]q=8.85\times 10^{-12}\ F/m\times 7.5\times 10^5\ V/m\times 0.0045\ m^2[/tex]

[tex]q=2.98\times 10^{-8}\ C[/tex]

So, the charge on the each plate is [tex]2.98\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

To determine the charge on each plate for a desired electric field in a capacitor with parallel plates, use the formula Q = E x A x ε_0. Convert the area to m^2, insert all values including the permittivity of free space, and solve for Q.

To find the charge required on each of the parallel plates to create a specific electric field, you can use the relationship between the electric field (E), the charge (Q) on the plates, and the permittivity of free space (ε0). The electric field between two parallel plates is given by the equation:

E = Q / (ε0 × A)

where E is the electric field, Q is the charge on each plate, A is the area of the plates, and ε0 is the permittivity of free space (ε0 = 8.854 x 10−12 C2/N·m2). Rearranging this for Q gives:

Q = E × A × ε0

First, convert the area from cm2 to m2 by multiplying by (10−4)2. Then, plug in the values:

Q = (7.50 × 105 V/m) × (0.45 × 10−4 m2) × (8.854 × 10−12 C2/N·m2)

Calculate Q to find the charge required on each plate.

D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.

In order to solve this problem we have to use the equation that relates resistance and resistivity:

R = ρL/A

Where ρ is the resistivity of the matter, the length of the wire, and A the area of ​​the cross section of the wire.

If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.

First, we have to clear A from the equation R = ρL/A:

A = ρL/R

Substituting the values

A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)

A = 1.9x10⁻⁷m²

The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:

D = √4A/π

Substituting the value of A:

D = √4(1.9x10⁻⁷m²)/π

D = 497.4x10⁻⁶m

The diameter of the wire is approximately 4.5 × 10^-6 meters.

To find the diameter of the wire, we can use the formula for resistance:

R = (ρ * L) / (A),

where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. Rearranging the formula to solve for the area:

A = (ρ * L) / R.

Given that the mile of 24-gauge copper wire has a resistance of 0.14 kΩ (or 0.14 * 10^3 Ω), the resistivity of copper is 1.7 × 10^-8 Ω ⋅ m, and 1 mile is equal to 1.6 km, we can substitute the values in the formula:

A = (1.7 × 10^-8 Ω ⋅ m * (1.6 * 10^3 m)) / (0.14 * 10^3 Ω).

Calculating the area, we find:

A ≈ 1.94 × 10^-11 m^2.

Next, we use the formula for the area of a circle (A = π * r^2) to find the radius (r) of the wire:

r = √(A / π).

Substituting the values, we have:

r ≈ √(1.94 × 10^-11 m^2 / π).

Calculating the radius, we find:

r ≈ 2.25 × 10^-6 m.

Finally, we can double the radius to find the diameter of the wire:

D = 2r ≈ 2 * 2.25 × 10^-6 m = 4.5 × 10^-6 m.

Therefore, the diameter of the wire is approximately 4.5 × 10^-6 meters.

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Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

Answer:

(a) 250 kg/m^3

(b) 2 m^3

(c) 1000 kg

Explanation:

(a) Let V be the total volume of the boat.

Volume immersed = 25 % of total volume = 25 V / 100   = V / 4

Let the density of boat is d.

Density of water = 1000 kg/m^3

Use the principle of flotation

Buoyant force on the boat = weight of the boat

Volume immersed x density of water x g = Volume x density of boat x g

V / 4 x 1000 x g = V x d x g

d = 250 kg/m^3

(b)

mass = 500 kg

density = 250 kg/m^3

V = mass / density = m / d = 500 / 250 = 2 m^3

(c) Let the mass of cargo is m.

Volume immeresed = 75 % of total volume = 75 x 2 / 100 = 1.5 m^3

Buoyant force = weight of boat + weight of cargo

1.5 x 1000 x g = 500 g + m g

1500 = 500 + m

m = 1000 kg

Answer:

Work done, W = 39.2 J

Explanation:

It is given that,

Mass of the block, m = 2 kg

The block is lifted vertically 2 m by the man i.e the distance covered by the block is, h = 2 m. The man is doing work against the gravity. It is given by :

[tex]W=mgh[/tex]

Where

g is acceleration due to gravity

[tex]W=2\ kg\times 9.8\ m/s^2\times 2\ m[/tex]

W = 39.2 J

So, the work done by the man is 39.2 J. Hence, this is the required solution.

Answer:

[tex]P = 2.94 \times 10^5 Pa[/tex]

Explanation:

Normal force due to four tires is counter balancing the weight of the car

So here we will have

[tex]4F_n = mg[/tex]

[tex]F_n = \frac{mg}{4}[/tex]

[tex]F_n = \frac{1.20 \times 10^3 \times 9.81}{4}[/tex]

[tex]F_n = 2943 N[/tex]

now we know that pressure in each tire is given by

[tex]P = \frac{F}{A}[/tex]

Here we know that

[tex]A = 1.00 \times 10^2 cm^2 = 1.00 \times 10^{-2} m^2[/tex]

[tex]P = \frac{2943}{1.00 \times 10^{-2}}[/tex]

[tex]P = 2.94 \times 10^5 Pa[/tex]

Answer:

P = 294300Pa or 42.67psi by conversion.

Explanation:

Since Four tyres were inflated, we have that area of the four tyres are

4×1×10²cm²

Pressure is given as:

P = f/a but f = mg

P = m×g/a

Therefore,

P = 1.20x10³kg × 9.81m/s² / (4 ×1x10² cm²)

P = 1.20x10³kg×9.81m/s² / (0.04m²)

P = 294300Pa or 42.67psi by conversion.

Answer:

0.48

Explanation:

m = mass of disc shaped grindstone = 50 kg

d = diameter of the grindstone = 0.52 m

r = radius of the grindstone = (0.5) d = (0.5) (0.52) = 0.26 m

w₀ = initial angular speed = 850 rev/min = 850 (0.10472) rad/s = 88.97 rad/s

t = time taken to stop = 7.5 sec

w = final angular speed  0 rad/s

Using the equation

w = w₀ + α t

0 = 88.97 + α (7.5)

α = - 11.86 rad/s²

F = normal force on the disc by the ax = 160 N

μ = Coefficient of friction

f = frictional force

frictional force is given as

f = μ F

f = 160 μ

Moment of inertia of grindstone is given as

I = (0.5) m r² = (0.5) (50) (0.26)² = 1.69 kgm²

Torque equation is given as

r f = I |α|

(0.26) (160 μ) = (1.69) (11.86)

μ = 0.48

Answer:

The drift velocity of electrons in a copper wire is [tex]1.756\times10^{-4}\ m/s[/tex]

Explanation:

Given that,

Density [tex]\rho=8.93\ g/cm^3[/tex]

Mass [tex]M=63.5\ g/mol[/tex]

Radius = 0.625 mm

Current = 3 A

We need to calculate the drift velocity

Using formula of drift velocity

[tex]v_{d}=\dfrac{J}{ne}[/tex]

Where, n = number of electron

j = current density

We need to calculate the current density

Using formula of current density

[tex] J=\dfrac{I}{\pi r^2}[/tex]

[tex] J=\dfrac{3}{3.14\times(0.625\times10^{-3})^2}[/tex]

[tex] J=2.45\times10^{6}\ A/m^2[/tex]

Now, we calculate the number of electron

Using formula of number of electron

[tex]n=\dfrac{\rho}{M}N_{A}[/tex]

[tex]n=\dfrac{8.93\times10^{6}}{63.5}\times6.2\times10^{23}[/tex]

[tex]n=8.719\times10^{28}\ electron/m^3[/tex]

Now put the value of n and current density into the formula of drift velocity

[tex]v_{d}=\dfrac{2.45\times10^{6}}{8.719\times10^{28}\times1.6\times10^{-19}}[/tex]

[tex]v_{d}=1.756\times10^{-4}\ m/s[/tex]

Hence, The drift velocity of electrons in a copper wire is [tex]1.756\times10^{-4}\ m/s[/tex]

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

Final answer:

The maximum compression of the spring, when a block of mass 1.5 kg is released from rest at a height of 0.81 m on a frictionless ramp onto a spring with spring constant 250.0 N/m, is approximately 30.7 cm.

Explanation:

To determine the maximum compression of the spring, x, we use the conservation of energy principle. The gravitational potential energy (PE) at the height H will convert to the elastic potential energy (spring PE) in the spring when the block of mass m compresses the spring at the foot of the ramp.

Initially, the block has gravitational potential energy given by PE = mgh, where g = 9.81 m/s2 is the acceleration due to gravity. At the point of maximum compression, all this energy will be stored as elastic potential energy in the spring given by spring PE = ½kx2.

Setting the initial PE equal to the spring PE and solving for x, we get:

mgh = ½kx2

1.5 kg × 9.81 m/s2 × 0.81 m = ½ × 250.0 N/m × x2

x2 = (1.5 kg × 9.81 m/s2 × 0.81 m) / (0.5 × 250.0 N/m)

x2 = (11.7915 kg·m2/s2) / (125.0 N/m)

x2 = 0.094332 m2

x = √0.094332 m2

x ≈ 0.307 m or 30.7 cm

Therefore, the maximum compression of the spring is approximately 30.7 cm.

Answer:

(a) 4323.67 N

(b) 421.13 degree

Explanation:

A = 2580 N 36 degree South of east

B = 2270 N due South

A = 2580 ( Cos 36 i - Sin 36 j) = 2087.26 i - 1516.49 j

B = 2270 (-j) = - 2270 j

A + B = 2087.26 i - 1516.49 j - 2270 j

A + B = 2087.26 i - 3786.49 j

(a) Magnitude of A + B = [tex]\sqrt{2087.26^{2}+(-3786.49)^{2}}[/tex]

Magnitude of A + B = 4323.67 N

(b) Direction of A + B

Tan theta = - 3786.49 / 2087.26

theta = - 61.13 degree

Angle from ast = 360 - 61.13 = 421.13 degree.

To determine the magnitude and direction of the resultant force acting on a tree stump when two forces are applied, one must resolve the forces into components, sum the components, then apply the Pythagorean theorem for magnitude and inverse tangent for direction.

We need to resolve each force into its horizontal (east-west) and vertical (north-south) components. For Force A, we can use trigonometry to find the components:

Force B is already pointing due south, so its components are:

The net force components are:

With these components, the magnitude of the resultant force can be calculated using the Pythagorean theorem:

Fnet = √(Fnet,x² + Fnet,y²)

To find the direction of the resultant force relative to due east, we can use the inverse tangent function (tan-1):

θ = tan-1(Fnet,y / Fnet,x)

This angle should be positive, as we are measuring it with respect to due east.

Answer:

Explanation:

Force, F = - mg j

r = - 7x i + y j

Torque is defined as the product f force and the perpendicular distance.

It is also defined as the cross product of force vector and the displacement vector.

[tex]\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}[/tex]

[tex]\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)[/tex]

[tex]\overrightarrow{\tau  }= 7 m g x k

Here, we observe that the torque is independent of y coordinate.

Answer:

i = 0.2 A

Explanation:

R = resistance of the resistor = 1000 Ω

V = Potential difference across the capacitor = Potential difference across the resistor = 200 V

i = initial current just after the capacitor is connected to resistor

Using ohm's law

[tex]i = \frac{V}{R}[/tex]

Inserting the values

[tex]i = \frac{200}{1000}[/tex]

i = 0.2 A

The initial current flowing through a 1000-Ω resistor when a 20.0-μF capacitor charged to 200 V is connected to it is 200 mA.

To find the initial current, we use Ohm's law, which states I = V / R, where I is the current, V is the voltage, and R is the resistance. At the moment the capacitor is connected to the resistor, the voltage across the resistor is the same as the voltage on the capacitor, so we have I = 200 V / 1000 Ω. The initial current is therefore 0.2 A, or 200 mA.

Answer:

2.12 rpm

Explanation:

[tex]a_c= Centripetal\ acceleration=3.00\ m/s^2\\r=radius=\frac {120}{2}=60\ m\\a_c=\frac {v^2}{r}\\\Rightarrow v^2=a_c\times r\\\Rightarrow v^2=3\times 60=180\\\Rightarrow v=\sqrt{180}=13.41\ m/s\\[/tex]

[tex]v=\omega r\\\Rightarrow \omega=\frac {v}{r}\\\Rightarrow \omega=\frac {13.41}{60}\\\Rightarrow \omega=0.2236\ rad/s\\1\ rad/s=9.55\ rpm\\\Rightarrow 0.2236\ rad/s=2.12\ rpm\\\therefore Wheel\ rotations=2.12\ rpm[/tex]

Final answer:

To generate an artificial gravity of 3.00 m/s² on a space station with a 120 m diameter, it must rotate at approximately 2.14 revolutions per minute.

Explanation:

To find the rate of the wheel's rotation in revolutions per minute (rpm) that will provide an artificial gravity of 3.00 m/s² for a space station with a diameter of 120 m, we use the formula for centripetal acceleration:

a = ω² × r

where a is the centripetal acceleration, ω is the angular velocity in radians per second, and r is the radius of the circle.

Since we are looking for ω in revolutions per minute and not radians per second, we will need to convert our solution. First, we find ω in radians per second by rearranging the formula:

ω = √(a/r)

Given a = 3.00 m/s² and the radius r = 60 m (half the diameter),

ω = √(3.00 m/s² / 60 m) = √0.05 s²/m = 0.224 rad/s

One revolution is 2π radians and there are 60 seconds in a minute, so we can find the number of revolutions per minute by

rev/min = ω × (60 s/min) / (2π rad/rev)

Plugging in our value of ω:

rev/min = 0.224 rad/s × (60 s/min) / (2π rad/rev) ≈ 2.14 rev/min

Therefore, the space station must rotate at approximately 2.14 revolutions per minute to produce the desired artificial gravity.

Answer:

Heat generated in the rod is 8.33 watts.

Explanation:

It is given that,

Length of rod, l = 2 m

Area of cross section, [tex]A=2\ mm\times 2\ mm=4\ mm^2=4\times 10^{-6}\ m^2[/tex]

Resistivity of rod, [tex]\rho=6\times 10^{-8}\ \Omega-m[/tex]

Potential difference, V = 0.5 V

The value of resistance is given by :

[tex]R=\rho\dfrac{l}{A}[/tex]

[tex]R=6\times 10^{-8}\ \Omega-m\times \dfrac{2\ m}{4\times 10^{-6}\ m^2}[/tex]

R = 0.03 ohms

Let H is the rate at which heat is generated in the rod . It is given by :

[tex]\dfrac{H}{t}=I^2R[/tex]

Since, [tex]I=\dfrac{V}{R}[/tex]

[tex]\dfrac{H}{t}=\dfrac{V^2}{R}[/tex]

[tex]\dfrac{H}{t}=\dfrac{(0.5)^2}{0.03}[/tex]

[tex]\dfrac{H}{t}=8.33\ watts[/tex]

So, the at which heat is generated in the rod is 8.33 watts. Hence, this is the required solution.

Answer:

12 × 10^12 Nm^2/C

Explanation:

According to the Gauss's theorem

The total electric flux by the enclosed charge is given by

Charge enclosed / €0

So flux by each surface

= Q enclosed / 6 × €0

= 641 / ( 6 × 8.854 × 10^-12)

= 12 × 10^12 Nm^2/C

(a) 107.2 Ω

The voltage in the circuit is written in the form

[tex]V=V_0 sin(\omega t)[/tex]

where in this case

V0 = 35.0 V is the maximum voltage

[tex]\omega=100 rad/s[/tex]

is the angular frequency

Given the inductance: L = 165 mH = 16.5 H, the reactance of the inductance is:

[tex]X_L = \omega L=(100)(0.165)=16.5 \Omega[/tex]

Given the capacitance: [tex]C=99\mu F=99\cdot 10^{-6} F[/tex], the reactance of the capacitor is:

[tex]X_C = \frac{1}{\omega C}=\frac{1}{(100)(99\cdot 10^{-6})}=101.0 \Omega[/tex]

And given the resistance in the circuit, R = 66.0 Ω, we can now find the impedance of the circuit:

[tex]Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{66^2+(16.5-101.0)^2}=107.2\Omega[/tex]

(b) 0.33 A

The maximum current can be calculated by using Ohm's Law for RLC circuit. In fact, we know the maximum voltage:

V0 = 35.0 V

The equivalent of Ohm's law for an RLC circuit is

[tex]I=\frac{V}{Z}[/tex]

where

Z=107.2 Ω

is the impedance. Substituting these values into the formula, we find:

[tex]I=\frac{35.0}{107.2}=0.33 A[/tex]

(c) 100 rad/s

The equation for the current is

[tex]I=I_0 sin(\omega t-\Phi)[/tex]

where

I0 = 0.33 A is the maximum current, which we have calculated previously

[tex]\omega[/tex] is the angular frequency

[tex]\Phi[/tex] is the phase shift

In an RLC circuit, the voltage and the current have the same frequency. Therefore, we can say that

[tex]\omega=100 rad/s[/tex]

also for the current.

(d) -0.907 rad

Here we want to calculate the phase shift [tex]\Phi[/tex], which represents the phase shift of the current with respect to the voltage. This can be calculated by using the equation:

[tex]\Phi = tan^{-1}(\frac{X_L-X_C}{R})[/tex]

Substituting the values that we found in part a), we get

[tex]\Phi = tan^{-1}(\frac{16.5-101.0}{66})=-52^{\circ}[/tex]

And the sign is negative, since the capacitive reactance is larger than the inductive reactance in this case. Converting in radians,

[tex]\theta=-52 \cdot \frac{2\pi}{360}=-0.907 rad[/tex]

So the complete equation of the current would be

[tex]I=0.33sin(100t+0.907)[/tex]

The impedance of the circuit is 107.2 Ω.

The maximum current is 0.33 A.

The numerical value for ω in the equation i = Imax sin (ωt − ϕ). rad/s is 100 rad/s.

The numerical value for ϕ in the equation i = Imax sin (ωt − ϕ) is -0.907 rad.

1. To find the voltage, we already know that V0=35V

And the voltage in the circuit is V= V0sin(wt)

Hence, the angular frequency, w= 100 rad/s

Given the inductance is 16.5Ω and the capacitance is 101Ω.

And the resistance is 66Ω

Hence, the impedance of the circuit is [tex]\sqrt{66^{2} + (16.5-101)^2} =107.2[/tex]Ω

2. To find the maximum current, since Z= 107.2Ω,

Recall, I = V/Z

Put the values into the formula

I = 0.33A.

3. To find the numerical value for ω in the equation i = Imax sin (ωt − ϕ)
We recall that I0 = 0.33A

In an RLC circuit, the voltage and the current have the same frequency. Therefore, we can say that

w= 100 rad/s.

4. To find the numerical value for ϕ in the equation i = Imax sin (ωt − ϕ)

Because the sign is in the negative and since the capacitive reactance is larger than the inductive reactance in this case.

Converting in radians,

θ = -52. 2π/360

= -0.907rad.

Hence, the complete equation would be:

I = 0.33sin(100t + 0.907)

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The vertical height between the climbers can be calculated from the projectile motion's kinematic formula: h = (v₀² sin²θ) / 2g. Here, the first aid kit has reached its trajectory's peak when caught, and its vertical velocity component is zero, making it ideal to implement the equation.

In your question, you want to know the vertical height between the two climbers. This is a question in projectile motion in physics. When an object is thrown with an initial velocity at an angle, it creates a trajectory as it travels. When the climber above catches the first aid kit, it's mentioned that it is moving horizontally, which means its vertical speed is zero at that very moment. Hence the first aid kit has reached the maximum height of its trajectory or the peak where the vertical component of the velocity is zero.

We can calculate the vertical distance or height (h) using a formula from kinematics. The formula is: h = (v₀² sin²θ) / 2g. Where v₀ is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity (which is ~9.8 m/s² on Earth). Plugging in the values we get: h = (5.70 m/s)² × sin²(45.7 °) / (2 * 9.8 m/s²). Solve this to get the height.

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The vertical height between the two climbers is approximately 0.8616 meters.

The vertical height between the two climbers is given by the equation for the vertical displacement in projectile motion:

[tex]\[ h = \frac{{v_{0y}^2}}{{2g}} \][/tex]

First, we need to find the initial vertical component of the velocity[tex](\( v_{0y} \))[/tex]. Since the initial velocity[tex](\( v_0 \))[/tex] is given as 5.70 m/s at an angle of 45.7 we can use the sine function to find [tex]\( v_{0y} \):[/tex]

[tex]\[ v_{0y} = v_0 \sin(\theta) \][/tex]

[tex]\[ v_{0y} = 5.70 \, m/s \times \sin(45.7) \][/tex]

Now, we calculate the sine of 45.7 and multiply it by the initial velocity:

[tex]\[ v_{0y} = 5.70 \, m/s \times 0.721 \][/tex]

[tex]\[ v_{0y} = 4.1117 \, m/s \][/tex]

Next, we use the equation for vertical displacement to find the height h :

[tex]\[ h = \frac{{(4.1117 \, m/s)^2}}{{2 \times 9.81 \, m/s^2}} \][/tex]

[tex]\[ h = \frac{{16.9046 \, m^2/s^2}}{{19.62 \, m/s^2}} \][/tex]

[tex]\[ h = 0.8616 \, m \][/tex]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]

[tex]3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}[/tex]

[tex]-3i+2j=3.0\times v_{2}[/tex]

[tex]v_{2}=-i+0.66j[/tex]

The direction of the momentum

[tex]tan\theta=\dfrac{0.66}{-1}[/tex]

[tex]\theta=tan^{-1}\dfrac{0.66}{-1}[/tex]

[tex]\theta=-33.42^{\circ}[/tex]

The direction of the momentum with respect to east

[tex]\theta=180-33.42=146.58^{\circ}[/tex]

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Answer:

26.67 Volt

Explanation:

number of electrons, n = 5 x 10^21

R = 20 ohm

t = 10 min = 10 x 60 = 600 sec

V = ?

q = n x charge of one electron

q = n x e

q = 5 x 10^21 x 1.6 x 10^-19 = 800 C

i = q / t = 800 / 600 = 4 / 3 A

According to Ohm's law

V = i x R

V = 4 x 20 / 3 = 26.67 Volt

The voltage or potential difference across the resistor is 26.67 Volt.

Given the data in the question;

Ohm’s law states that the potential difference between two points is directly proportional to the current flowing through the resistance.

Hence

[tex]V = IR[/tex]

Where V is the voltage or potential difference, potential difference, I is the current and R is the resistance.

First we determine the total charge.

[tex]Q = n * e[/tex]

Where Q is the charge, n is the number of electrons and e is the charge on one electron. ( [tex]e = 1.6 * 10^{-19}C[/tex] )

Hence

[tex]Q = (5*10^{21}) * (1.6 * 10^{-19C)}\\\\Q = 800C[/tex]

Since current is the electric charge transferred per unit time.

[tex]I = \frac{Q}{t}[/tex]

Hence

[tex]I = \frac{800C}{600s}\\ \\I = 1.33A[/tex]

Now, we input our values into the above expression

[tex]V = I * R\\\\V = 1.33A * 20Ohms\\\\V = 26.67Volt[/tex]

Therefore, the voltage or potential difference across the resistor is 26.67 Volt.

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Answer:

New speed of water is 2.8 m/s.

Explanation:

It is given that,

Speed of water, v₁ = 2.3 m/s

Diameter of pipe, d₁ = 3.2 cm

Radius of pipe, r₁ = 1.6 cm = 0.016 m

Area of pipe, [tex]A_1=\pi(0.016)^2=0.000804\ m^2[/tex]

If the pipe narrows its diameter, d₂ = 2.9 cm

Radius, r₂ = 0.0145 m

Area of pipe, [tex]A_2=\pi(0.0145)^2=0.00066\ m^2[/tex]

We need to find the new speed of the water. It can be calculated using equation of continuity as :

[tex]v_1A_1=v_2A_2[/tex]

[tex]v_2=\dfrac{v_1A_1}{A_2}[/tex]

[tex]v_2=\dfrac{2.3\ m\times 0.000804\ m^2}{0.00066\ m^2}[/tex]

[tex]v_2=2.8\ m/s[/tex]

So, the new speed of the water is 2.8 m/s. Hence, this is the required solution.

Answer:

v/U=0.79

Explanation:

Given u=[tex]u=\frac{Uy}{\partial }=\frac{Uy}{cx^{1/2}}[/tex]

Now for the given flow to be possible it should satisfy continuity equation

[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0[/tex]

Applying values in this equation we have

[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\\\\\frac{\partial u}{\partial x}=\frac{\partial (\frac{Uy}{cx^{1/2}})}{\partial x}\\\\\frac{\partial u}{\partial x}=\frac{-1}{2}\frac{Uy}{cx^{3/2}}\\\\[/tex]

Thus we have

[tex]\frac{\partial v}{\partial y}=\frac{1}{2}\frac{Uy}{cx^{3/2}}\\\\\therefore \int \partial v=\int \frac{1}{2}\frac{Uy}{cx^{3/2}}\partial x\\\\v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}\\\\v=\frac{1}{4}\frac{uy}{x}[/tex] Hence proved [tex]\because u=\frac{Uy}{cx^{1/2}}[/tex]

For maximum value of v/U put y =[tex]\partial[/tex]

[tex]v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}[/tex]

[tex]v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}\\\\\frac{v}{U}=\frac{\partial ^{2}}{4cx^{3/2}}\\\\[/tex]

Thus solving we get using the given values

v/U=0.79

The y-component of velocity in a boundary layer flow is derived using the principle of continuity for an incompressible fluid. Upon evaluation, the ratio v /U has a maximum of 0.25 at the specified location (x = 0.5 m and δ = 5 mm).

The y-component v of the velocity can be solved using the continuity equation, d/dx (u A) + d/dy (v A) = 0, where A is the cross-sectional area of the pipe. This equation arises from the principle of conservation of mass for an incompressible fluid. In this case, our A is the width of the plate (into the page) times the y-distance from the plate or y*b. When the equation derived for velocity, u = Uy/δ, and the equation for the boundary layer thickness, δ = cx1/2, are plugged into the continuity equation and simplified, we derive the expression v = uy/4x.  Substituting for δ, we get v = Uy/(4δ*sqrt(x)), and at x = 0.5 m and δ = 5 mm, the maximum value of v /U is 1/4 or 0.25.

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Answer:

The tension of the rope is T= 172.52 N

Explanation:

m= 21 kg

α= 21º

μ= 0.285

a= 2.09 m/s²

g= 9.81 m/s²

W= m*g

W=206.01 N

Fr= μ*W*cos(α)

Fr= 54.81 N

Fx= W * sin(α)

Fx= 73.82 N

T-Fr-Fx = m*a

T= m*a + Fr + Fx

T= 172.52 N

The magnitude of the force of motion of an object pulled by a rope up an inclined plane is the tension in the rope less the frictional forces as well as the component of the weight of the object acting along the plane

The tension in the rope is approximately 172.53 Newtons

Known:

The mass of the box, m = 21.0 kg

The angle of inclination of the ramp, θ = 21.0°

The coefficient of kinetic friction, μk= 0.285

The acceleration of the box up the ramp, a = 2.09 m/s²

The acceleration due to gravity, g = 9.81 m/s²

Required:

The tension in the rope, T

Solution;

The normal reaction of the box, N = m × g × cos(θ)

∴ N = 21.0 × 9.81 × cos(21.0°) ≈ 192.33

The normal reaction, N ≈ 192.33 N

The frictional force, [tex]F_f[/tex] = μ × N

∴ [tex]F_f[/tex] = 0.285 × 192.33 ≈ 54.81405

The frictional force, [tex]F_f[/tex] ≈ 54.81405 N

The force pulling the box, F = m×a = T - [tex]F_f[/tex] - The component of the weight acting along the plane

The component of the weight acting along the plane = m·g·sin(θ)

∴ m×a = T - [tex]F_f[/tex] - m·g·sin(θ)

T = m×a + [tex]F_f[/tex] + m·g·sin(θ)

Which gives;

T = 21.0 × 2.09 + 54.81405 + 21.0 × 9.81 × sin(21.0°) ≈ 172.53

The tension in the rope, T ≈ 172.53 N

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Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

The final image position is 4.34 cm to the right of the second lens, and the final image height is 2.4 cm.

First, we will consider the object in front of the first lens (with a focal length of 40 cm). The object is placed 14 cm in front of this lens.

For the first lens, the lens equation is given by:

[tex]\[\frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_i}\][/tex]

where [tex]\(f_1\)[/tex] is the focal length of the first lens, [tex]\(d_o\)[/tex] is the object distance from the first lens, and [tex]\(d_i\)[/tex] is the image distance from the first lens.

Rearranging the equation to solve for [tex]\(d_i\)[/tex], we get:

[tex]\[d_i = \frac{f_1 \cdot d_o}{d_o - f_1}\][/tex]

Plugging in the values:

[tex]\[d_i = \frac{40 \cdot 14}{14 - 40}\] \[d_i = \frac{560}{-26}\] \[d_i = -21.54 \text{ cm}\][/tex]

This means the image formed by the first lens is 21.54 cm to the left of the first lens (virtual image).

Next, we calculate the height of this image using the magnification equation:

[tex]\[m_1 = -\frac{d_i}{d_o}\] \[m_1 = -\frac{-21.54}{14}\] \[m_1 = 1.54\][/tex]

The height of the image formed by the first lens is:

[tex]\[h_i = m_1 \cdot h_o\][/tex]

where [tex]\(h_o\)[/tex] is the object height (2.0 cm).

So, [tex]\(h_i = 1.54 \cdot 2.0\)[/tex]

[tex]\[h_i = 3.08 \text{ cm}\][/tex]

Now, this image becomes the object for the second lens (with a focal length of 20 cm). The distance between the two lenses is 16 cm, so the object distance for the second lens is the image distance from the first lens plus 16 cm.

For the second lens, the object distance [tex]\(d_o'\)[/tex] is:

[tex]\[d_o' = -21.54 + 16\] \[d_o' = -5.54 \text{ cm}\][/tex]

Using the lens equation for the second lens:

[tex]\[\frac{1}{f_2} = \frac{1}{d_o'} + \frac{1}{d_i'}\][/tex]

where [tex]\(f_2\)[/tex] is the focal length of the second lens, [tex]\(d_o'\)[/tex] is the object distance from the second lens, and [tex]\(d_i'\)[/tex] is the image distance from the second lens.

Rearranging the equation to solve for [tex]\(d_i'\)[/tex], we get:

[tex]\[d_i' = \frac{f_2 \cdot d_o'}{d_o' - f_2}\][/tex]

Plugging in the values:

[tex]\[d_i' = \frac{20 \cdot (-5.54)}{-5.54 - 20}\] \[d_i' = \frac{-110.8}{-25.54}\] \[d_i' = 4.34 \text{ cm}\][/tex]

This means the final image is 4.34 cm to the right of the second lens (real image).

To find the height of the final image, we use the magnification equation for the second lens:

[tex]\[m_2 = -\frac{d_i'}{d_o'}\] \[m_2 = -\frac{4.34}{-5.54}\] \[m_2 = 0.78\][/tex]

The height of the final image is:

[tex]\[h_i' = m_2 \cdot h_i\] \[h_i' = 0.78 \cdot 3.08\] \[h_i' = 2.4 \text{ cm}\][/tex]

Answer:

[tex]K.E =0.081eV[/tex]

[tex]\vec{\Omega}=(0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]

Explanation:

Given:

Velocity vector [tex]\vec{v}=(2132\hat{i}+3300\hat{j}+154\hat{k})m/s[/tex]

the mass of neutron, m = 1.67 × 10⁻²⁷ kg

Now,

the kinetic energy (K.E) is given as:

[tex]K.E =\frac{1}{2}mv^2[/tex]

[tex]\vec{v}^2 = \vec{v}.\vec{v}[/tex]

or

[tex]\vec{v}^2 = (2132\hat{i}+3300\hat{j}+154\hat{k}).(2132\hat{i}+3300\hat{j}+154\hat{k})[/tex]

or

[tex]\vec{v}^2 =15.45\times 10^6 m^2/s^2[/tex]

substituting the values in the K.E equation

[tex]K.E =\frac{1}{2}1.67\times 10^{-27}kg\times 15.45\times 10^6 m^2/s^2[/tex]

or

[tex]K.E =1.2989\times 10^{-20}J[/tex]

also

1J = 6.242 × 10¹⁸ eV

thus,

[tex]K.E =1.2989\times 10^{-20}\times 6.242\times 10^{18}[/tex]

[tex]K.E =0.081eV[/tex]

Now, the direction vector [tex]\vec{\Omega}[/tex]

[tex]\vec{\Omega}=\frac{\vec{v}}{\left | \vec{v} \right |}[/tex]

or

[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{\left |\sqrt{((2132\hat{i})^2+(3300\hat{j})^2+(154\hat{k}))^2} \right |}[/tex]

or

[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{3930.65}[/tex]

or

[tex]\vec{\Omega}=0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]

Explanation:

We know

the frequency of the tube with one end open and the other end closed follows the given relations as

[tex]f_{1}[/tex] : [tex]f_{2}[/tex] : [tex]f_{3}[/tex] : [tex]f_{4}[/tex] = 1 : 3 : 5 : 7

∴ the 4th allowed wave is [tex]f_{4}[/tex] = 7 [tex]f_{1}[/tex]

                                                                   = [tex]\frac{7v}{4l}[/tex]

We know  [tex]f_{4}[/tex] = 1975 Hz and v = 343 m/s ( as given in question )

∴[tex]l = \frac{7\times v}{4\times f_{4}}[/tex]

 [tex]l = \frac{7\times 343}{4\times 1975}[/tex]

           = 0.303 m

We know that v = [tex]f_{4}[/tex] x [tex]λ_{4}[/tex]

[tex]\lambda _{4}= \frac{v}{f_{4}}[/tex]

[tex]\lambda _{4}= \frac{343}{1975}[/tex]

                            = 0.17 m

Now when the warmer air is flowing, the speed gets doubled and the mean temperature increases. And as a result the wavelength increases but the amplitude and the frequency remains the same.

So we can write

v ∝ λ

or  [tex]\frac{v_{1}}{v_{2}}= \frac{\lambda _{1}}{\lambda _{2}}[/tex]

Therefore, the wavelength becomes doubled = 0.17 x 2

                                                                             = 0.34 m

Now the new length of the air column becomes doubled

∴ [tex]l^{'}[/tex] = 0.3 x 2

                        = 0.6 m

∴ New speed, [tex]v^{'}[/tex] = 2 x 343

                                               = 686 m/s

∴ New frequency is [tex]f^{'}=\frac{v^{'}}{4\times l^{'}}[/tex]

                                 [tex]f^{'}=\frac{686}{4\times 0.6}[/tex]

                                               = 283 Hz

∴ The new frequency remains the same.

Now we know

[tex]v_{s}[/tex] = 12 m/s, [tex]v_{o}[/tex] = 4 m/s, [tex]f_{o}[/tex] = 1975 Hz

Therefore, apparent frequency is [tex]f^{'}=f^{o}\left ( \frac{v+v_{s}}{v+v_{o}} \right )[/tex]

[tex]f^{'}=1975\left ( \frac{343+12}{343+4} \right )[/tex]

            = 2020.5 Hz