Two slits are illuminated by a 602 nm light. The angle between the zeroth-order bright band at the center of the screen and the fourth-order bright band is 17.4 ◦ . If the screen is 138 cm from the double-slit, how far apart is this bright?

Answer:

Rate of heat is 45.1 kJ/min

Explanation:

Heat required to evaporate the water is given by

Q = mL

here we know that

[tex]L = 2.25 \times 10^6 J/kg[/tex]

now we have

[tex]Q = (0.200)(2.25 \times 10^6 J/kg)[/tex]

[tex]Q = 452.1 kJ[/tex]

now the power is defined as rate of energy

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

[tex]P = \frac{452.1 kJ}{10}[/tex]

[tex]P = 45.1 kJ/min[/tex]

Answer:

[tex]\mu_k = 0.15[/tex]

Explanation:

according to the kinematic equation

[tex]v^{2} - u^{2} = 2aS[/tex]

Where

u is initial velocity  = 0 m/s

a = acceleration

S is distance = 8.00 m

final velocity = 1.0 m/s

[tex]a = \frac {v^{2}}{2S}[/tex]

[tex]a = \frac {1{2}}{2*8.6}[/tex]

a = 0.058 m/s^2

from newton second law

Net force = ma

[tex]f_{net} = ma[/tex]

F - f = ma

2[tex]5 - \mu_kN = ma[/tex]

[tex]25 - \mu_kmg = ma[/tex]

[tex]\frac {25 - ma}{mg} =\mu_k[/tex]

[tex]\frac {25 - 16*0.058}{16*9.81} = 0.15[/tex]

[tex]\mu_k = 0.15[/tex]

Answer:

a) 7250.5 N

b) 4.6 m/s²

Explanation:

a)

F = applied force = 8000 N

θ = angle with the horizontal = 65 deg

Consider the motion along the vertical direction :

[tex]F_{y}[/tex] = Applied force in vertical direction in upward direction = F Sinθ = 8000 Sin65 = 7250.5 N

[tex]F_{g}[/tex] = weight of the plane in vertical direction in downward direction = ?

[tex]a_{y}[/tex] = Acceleration in vertical direction = 0 m/s²

Taking the force in upward direction as positive and in downward direction as negative, the force equation along the vertical direction can be written as

[tex]F_{y}-F_{g} = m a_{y}[/tex]

[tex]7250.5 -F_{g} = m (0)[/tex]

[tex]F_{g}[/tex] = 7250.5 N

b)

m = mass of the plane

force of gravity is given as

[tex]F_{g} = mg [/tex]

[tex]7250.5 = m(9.8) [/tex]

m = 739.85 kg

Consider the motion along the horizontal direction

[tex]F_{x}[/tex] = Applied force in horizontal direction = F Cosθ = 8000 Cos65 = 3381 N

[tex]a_{x}[/tex] = Acceleration in horizontal direction

Acceleration in horizontal direction is given as

[tex]a_{x}=\frac{F_{x}}{m}[/tex]

[tex]a_{x}=\frac{3381}{739.85}[/tex]

[tex]a_{x}[/tex] = 4.6 m/s²

Answer:It is rising at a rate of [tex]7.5cm/min[/tex]

Explanation:

We have volume of trapezoid equals

[tex]V=Area\times Length\\\\V=\frac{1}{2}(a+b)h\times L[/tex]

Thus at any time 't' we have

[tex]V(t)=\frac{1}{2}(a(t)+b(t))h(t)\times L\\\\\therefore V(t)=\frac{1}{2}(20+b(t))\times h(t)\times L[/tex]

Differentiating both sides with respect to time we get

[tex]\frac{dV(t)}{dt}=\frac{1}{2}b'(t)h(t)L+\frac{1}{2}(20+b(t))\times h'(t)L[/tex]

Applying values we have

[tex]b(t)=20+h(t)\\b'(t)=h'(t)[/tex]

Thus we have

[tex]\frac{dV(t)}{dt}=\frac{1}{2}h'(t)h(t)L+\frac{1}{2}(20+20+h(t))\times h'(t)L\\\\2V'(t)=h'(t)L[h(t)+(40+h(t))]\\\\\therefore h'(t)=\frac{2V'(t)}{L(h(t)+(40+h(t)))}[/tex]

Applying values we get

[tex]h'(t)=0.075m/min=7.5cm/min[/tex]

Answer:

275.4 kg

Explanation:

specific gravity = 0.918

Density = 0.918 g/cm^3 = 918 kg/m^3

Volume = 100 gallons = 300 litres = 300 x 10^-3 m^3 = 0.3 m^3

Mass = volume x density = 0.3 x 918 = 275.4 kg

Answer:

1.176m

Explanation:

Local ambient pressure(P1) = 100 kPa

Absolute pressure(P2)=260kPa

Net pressure=absolute pressure-local ambient absolute pressure

Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)

Net pressure=260-100=160kPa

Pressure= ρgh

160kPa=13600*10*h

h=[tex]\frac{160000}{136000}[/tex]

h=1.176m

From conservation of linear momentum, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.

There are four types of collision

In elastic collision, both momentum and energy are conserved. While in inelastic collision, only momentum is conserved.

From the given question, the following parameters are given.

Since the collision is inelastic, they will both move with a common velocity after collision.

Horizontal component

[tex]m_{1}[/tex][tex]v_{1}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex] ) V

1500 x 25 = (1500 + 2500) V

37500 = 4000V

V = 37500 / 4000

V = 9.375 m/s

Vertical component

[tex]m_{2}[/tex][tex]v_{2}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex])V

2500 x 20 = (1500 + 2500)V

50000 = 4000V

V = 50000 / 4000

V = 12.5 m/s

The net velocity will be the magnitude of the velocity of the wreckage after collision

V = [tex]\sqrt{9.4^{2} + 12.5^{2} }[/tex]

V = [tex]\sqrt{244.61}[/tex]

V = 15.6 m/s

The direction will be

Tan Ф = 12.5 / 9.4

Ф = [tex]Tan^{-1}[/tex](1.329)

Ф = 53 degrees.

Therefore,  the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.

Learn more about collision here: https://brainly.com/question/7694106

The velocity of the wreckage after the collision is 32.015 m/s in the direction 38.66° north of east.

This problem is related to the conservation of linear momentum which states the total momentum of an isolated system remains constant if no external forces act on it. Momentum is a vector quantity having both magnitude and direction. The total initial and final momentum in both horizontal (x) and vertical (y) directions must be equal.

Initial momentum of the car is (1500 kg)*(25.0 m/s) = 37500 kg*m/s in the east direction whereas initial momentum of the van is (2500 kg)*(20.0 m/s) = 50000 kg*m/s in the north direction.

Since the collision is perfectly inelastic, the two vehicles stick together after the collision and move as one. Therefore, the final momentum is the vector sum of the individual momenta. We therefore calculate the magnitude of the resultant velocity using Pythagoras theorem, √[(25.0 m/s)² + (20.0 m/s)²] = 32.015 m/s.

To find the direction of the final velocity, we use the tangent of the angle which is equal to the vertical component divided by the horizontal component, which gives us an angle of 38.66° north of east.

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Answer:

Part a)

the two frequencies are in ratio of odd numbers so it must be closed at one end

Part b)

L = 34.3 cm

Explanation:

Part a)

Since the shortest frequency is known as fundamental frequency

It is given as

[tex]f_o = 250 Hz[/tex]

next higher frequency is given as

[tex]f_1 = 750 Hz[/tex]

since the two frequencies here are in ratio of

[tex]\frac{f_1}{f_o} = \frac{750}{250} = 3 : 1[/tex]

since the two frequencies are in ratio of odd numbers so it must be closed at one end

Part b)

For the length of the pipe we can say that fundamental frequency is given as

[tex]f_o = \frac{v}{4L}[/tex]

here we have

[tex]250 = \frac{343}{4(L)}[/tex]

now we will have

[tex]L = \frac{343}{4\times 250}[/tex]

[tex]L = 34.3 cm[/tex]

Answer:

The acceleration of the ball is 666.67 m/s²

Explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Applied force to it, F = 100 N

We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :

F = ma

[tex]a=\dfrac{F}{m}[/tex]

[tex]a=\dfrac{100\ N}{0.15\ kg}[/tex]

[tex]a=666.67\ m/s^2[/tex]

So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.

Answer:

(a) 6.22 x 10^-6 C

(b) 3.8 x 10^13

Explanation:

Let the second charge is q2 = q

q1 = 1.6 x 10^-19 C

F = 3.5 x 10^9 N

d = 1.6 mm = 1.6 x 10^-3 m

(a) Use the formula of Coulomb's law

F = K q1 x q2 / d^2

3.5 x 10^-9 = 9 x 10^9 x 1.6 x 10^-19 x q / (1.6 x 10^-3)^2

q = 6.22 x 10^-6 C

(b)

Let the number of electrons be n

n = total charge / charge of one electron

n = 6.22 x 10^-6 / (1.6 x 10^-19) = 3.8 x 10^13

Answer:

The acceleration of the crate is 1 m/s²

Explanation:

It is given that,

Mass of the crate, m = 5 kg

Two forces applied on the crate i.e. one is 3.0 N to the north and the other is 4.0 N to the east. So, there resultant force is :

[tex]F_{net}=\sqrt{3^2+4^2} =5\ N[/tex]

We need to find the acceleration of the crate. It is given by using the second law of motion as :

[tex]a=\dfrac{F_{net}}{m}[/tex]

[tex]a=\dfrac{5\ N}{5\ kg}[/tex]

a = 1 m/s²

So, the acceleration of the crate is 1 m/s². Hence, this is the required solution.  

The magnitude of the acceleration of a crate with forces of 3.0 N north and 4.0 N east applied to it is 1.0 m/s². This is found using the Pythagorean theorem to calculate the resultant force and Newton's Second Law to calculate acceleration.

The forces are 3.0 N to the north and 4.0 N to the east on a 5.0-kg crate. Since the forces are perpendicular, we can use the Pythagorean theorem to find the resultant force. The resultant force (Fr) is √(3.02 + 4.02) N, which is 5.0 N. According to Newton's Second Law, F = ma, hence acceleration (a) is Fr divided by the mass (m). Calculating acceleration: a = 5.0 N / 5.0 kg = 1.0 m/s2. Therefore, the correct answer is a. 1.0 m/s2.

Answer:

[tex]v = \sqrt{v_0^2 - \frac{2k}{3}(s^3 - s_0^3)}[/tex]

v = 9.67 m/s

Explanation:

As we know that acceleration is rate of change in velocity

so it is defined as

[tex]a = \frac{dv}{dt}[/tex]

[tex]a = v\frac{dv}{ds}[/tex]

here we know that

[tex]a = - ks^2 = v\frac{dv}{ds}[/tex]

now we have

[tex]vdv = - ks^2ds[/tex]

integrate both sides we have

[tex]\int vdv = -k \int s^2ds[/tex]

[tex]\frac{v^2}{2} - \frac{v_0^2}{2} = -k(\frac{s^3}{3} - \frac{s_0^3}{3})[/tex]

[tex]v^2 = v_0^2 - \frac{2k}{3}(s^3 - s_0^3)[/tex]

here we know that

[tex]v_0 = 10 m/s[/tex]

[tex]s_0 = 3 m[/tex]

[tex]v^2 = 10^2 - \frac{2(0.10)}{3}(5^3 - 3^3)[/tex]

[tex]v = 9.67 m/s[/tex]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

[tex]Y = \frac{FL}{A\Delta L}[/tex]

[tex]20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}[/tex]

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

Answer:

Part a)

[tex]Q = 2.47 \times 10^6[/tex]

Part b)

t = 4950.3 s

Explanation:

As we know that heat required to raise the temperature of container and water in it is given as

[tex]Q = m_1s_2\Delta T_1 + m_2s_2\Delta T_2[/tex]

here we know that

[tex]m_1 = 0.750[/tex]

[tex]s_1 = 900[/tex]

[tex]m_2 = 2.50 kg[/tex]

[tex]s_2 = 4186[/tex]

[tex]\Delta T_1 = \Delta T_2 = 100 - 30 = 70^oC[/tex]

now we have

[tex]Q_1 = 0.750(900)(70) + (2.5)(4186)(70) = 779800 J[/tex]

now heat require to boil the water

[tex]Q = mL[/tex]

here

m = 0.750 kg

[tex]L = 2.25 \times 10^6 J/kg[/tex]

now we have

[tex]Q_2 = 0.750(2.25 \times 10^6) = 1.7 \times 10^6 J[/tex]

Now total heat required is given as

[tex]Q = Q_1 + Q_2[/tex]

[tex]Q = 779800 + 1.7 \times 10^6 = 2.47 \times 10^6 J[/tex]

Part b)

Time taken to heat the water is given as

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

here we know that

power = 500 W

now we have

[tex]t = \frac{2.47 \times 10^6}{500} = 4950.3 s[/tex]

Answer:

37.86 kg

Explanation:

The weight of sign board is equally divided on each rope. It means the tension in all the ropes is equal to the weight of the sign board in equilibrium condition.

Tension in each rope = 53 N

Tension in 7 ropes = 7 x 53 N = 371 N

Thus, The weight of sign = 371 N

Now, weight = m g

where m is the mass of sign.

m = 371 / 9.8 = 37.86 kg

Answer:

Force on eardrum

       [tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]

Explanation:

Force = Pressure x Area

Pressure = hρg

Height, h = 3 m

ρ = 1000 kg/m³

g = 9.8 m/s²

Pressure = hρg = 3 x 1000 x 9.8 = 29400 N/m²

Area = 1 cm²

Force on eardrum

       [tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]

Answer:

b) Projectile MOTION

Explanation:

SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line

In this type of motion particle must be in straight line motion

So here we can say

a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM

b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM

d) Spring Motion : it is a straight line to and fro motion so it is also a SHM

So correct answer will be

b) Projectile MOTION

Final answer:

Projectile motion is not a simple harmonic motion because it does not meet the conditions for SHM.

Explanation:

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is proportional to the displacement. The three conditions that must be met to produce SHM are: a linear restoring force, a constant force constant, and no external damping forces. Based on these conditions, the answer to the question is (b) Projectile motion, as it does not meet the conditions for SHM. A projectile follows a parabolic path and does not have a linear restoring force.

Answer:

[tex]\omega _{f}=0.3107rad/sec[/tex]

partb) [tex]\omega _{f}=14.93rad/sec[/tex]

Explanation:

Since the system is isolated it's angular momentum shall be conserved

[tex]L_{system}=I_{system}\omega \\\\I_{system}=I_{rod}+2\times m\times r^{2}\\\\I_{system}=\frac{1}{12}ml^{2}+2\times 0.2\times (4.8\times 10^{-2})^{2}\\\\I_{system}=1.22845\times 10^{-3}kgm^{2}\\\\L_{system}=1.22845\times 10^{-3}kgm^{2}\times 3.66rad/sec[/tex]

[tex]\L_{system}=4.496\times 10^{-3}kgm^{2}[/tex]

Now the final angular momentum of the system

[tex]L_{fianl}=I_{final}\omega \\\\I_{final}=I_{rod}+2\times m\times r_{f}^{2}\\\\I_{final}=\frac{1}{12}ml^{2}+2\times 0.2\times (0.19)^{2}\\\\I_{final}=0.01447kgm^{2}\\\\L_{final}=0.01447kgm^{2}\times \omega _{final}rad/sec[/tex]

Thus equating initial and final angular momentum we solve for final angular velocity of rod as

[tex]\omega _{f}=\frac{4.496\times 10^{-3}}{0.01447}\\\\\omega _{f}=0.3107rad/sec[/tex]

part b)

When the rings leave the system we again conserve the angular momentum just before the rings leave the system and the instant when they just leave

[tex]\therefore 0.01447=\frac{1}{12}ml^{2}w\\\\\therefore w=\frac{4.582\times 10^{-3}}{3.068\times10^{-4} }\\\\w_{final}=14.93rad/sec[/tex]

Answer:

A)[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].

B)[tex]\omega_{f}=140\frac{rev}{min}[/tex].

Explanation:

A)

For this problem, we will use the conservation of angular momentum.

[tex]L_{0}=L_{f}\\[/tex].

In the beginning, we have that

[tex]L_{0}=I_{0}\omega_{0}\\[/tex]

where [tex]I_{0}[/tex] is the inertia moment of all the system at the starting position, this is the inertia moment of the rod plus the inertia moment of each ring ([tex]mr^{2}[/tex], with [tex]r[/tex] the distance from the ring to the fixed axis and, [tex]m[/tex] its mass) at the starting position and, [tex]\omega_{0}[/tex] is the initial angular velocity. So

[tex]L_{0}=(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}[/tex].

When the rings are at the ends of the rod the angular momentum becomes

[tex]L_{f}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex],

where [tex]r_{f}[/tex] is the distance from the fixed axis to the end of the rod (the final position of the rings).

Using conservation of angular momentum we get

[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex].

thus

[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2}+2mr_{f}^{2})}[/tex]

computing this last expresion we get

[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2+2(0.200)(0.19)^{2})}[/tex]

[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].

B)

Again we use the conservation of angular momentum. The initial angular momentum if the same as before. The final angular momentum will be

[tex]L_{f}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex],

this time we will not take into account the inertia moment of the rings because they are no longer part of the system (they leave the rod).

[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex].

thus

[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2})}[/tex]

computing this last expresion we get

[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2})[/tex]

[tex]\omega_{f}=140\frac{rev}{min}[/tex].

Answer: The mass of the object will be 11250 kg.

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given :

Density of the object= [tex]12500kg/m^3[/tex]

Mass of object = ?

Volume of the object = [tex]0.9m^3[/tex]

Putting in the values we get:

[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]

[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]

[tex]mass=11250kg[/tex]

Thus the mass of the object is 11250 kg.

Answer:

314 Joule

Explanation:

m = 10 kg, g = 10 m/s^2, d = 1 m, angle turn = 180 degree = π radian

work = torque x angle turn

torque = force x perpendicular distance

torque = m x g x d = 10 x 10 x 1 = 100 Nm

work = 100 x π

work = 100 x 3.14 = 314 Joule

Final answer:

The work done by a constant torque on a rigid body rotating through an angle of 180° can be found by multiplying the torque by the angle. The torque can be calculated using the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. Substituting the given values, we find that the work done by the torque on the rigid body is 18000 kg·m^2·rad/s^2.

Explanation:

The work done by a torque on a rigid body is given by the formula W = τθ, where τ is the torque and θ is the angle through which the body rotates. In this case, the torque is constant in both magnitude and direction, so we can use W = τθ. Given that the torque is constant and the body turns through an angle of 180°, we can calculate the work done as follows:

Since τ is constant, we can write W = τθ = τ (180° - 0°). The work done by the torque is equal to the torque multiplied by the change in angle. Substitute the given values into the formula: W = (τ) (180° - 0°) = (τ) (180°). The work done by the torque is equal to the torque multiplied by 180°.

To find the value of the torque, we need to use the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. In this case, the rigid body has a mass of 10 kg and a distance of 1 m from the pivot. The moment of inertia for a point mass rotating about a fixed axis is given by I = m(r^2), where m is the mass and r is the perpendicular distance from the axis. Substitute the given values into the formula: I = (10 kg)((1 m)^2) = 10 kg·m^2. Since α = a/r and the acceleration due to gravity is 10 m/s^2, we have α = (10 m/s^2)/(1 m) = 10 rad/s^2.

Substitute the values of τ and α into the equation τ = Iα: τ = (10 kg·m^2)(10 rad/s^2) = 100 kg·m^2·rad/s^2. Therefore, the torque is 100 kg·m^2·rad/s^2.

Finally, substitute the values of τ and θ into the equation W = τθ: W = (100 kg·m^2·rad/s^2 )(180°) = 18000 kg·m^2·rad/s^2.

Answer:

explained

Explanation:

When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.

Light below certain frequency will not cause any electron emission no matter how intense.

The intensity produces more electron but does not change the maximum kinetic energy of electrons.

Work function is independent of the intensity of light, because it is an intrinsic property of a material.

Answer:

anterior

anterior

Explanation:

In the given question is asked that

If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.

The current answer to the blanks will be

anterior

anterior

hope this helps any further query can be asked in comment section.

Answer:

The distance of the mosquito from the bat is 4.25 m.

Explanation:

Given that,

Speed of sound in air v= 340 m/s

Time t = 0.0250 second

Let d be the distance of the mosquito from the bat.

The distance traveled by the sound when the echo heard is 2d.

We need to calculate the distance

Using formula of  distance

[tex]v = \dfrac{2d}{t}[/tex]

Put the value into the formula

[tex]v = \dfrac{2d}{25\times10^{-3}}[/tex]

[tex]d =\dfrac{25\times10^{-3}\times340}{2}[/tex]

[tex]d=4.25\ m[/tex]

Hence, The distance of the mosquito from the bat is 4.25 m.

Utilizing the echo time of 0.0250 seconds and the sound speed of 340 m/s, the distance to the mosquito is calculated as 4.25 meters away from the bat.

Calculating the Distance to the Mosquito using Echo Time

To determine the distance to the mosquito that reflects the bat's shriek, we need to use the speed of sound in air and the echo time. Since the shriek's echo returns in 0.0250 seconds and the speed of sound is given as 340 m/s, we can calculate the total distance traveled by the sound (to the mosquito and back to the bat) with the equation:

Distance = Speed × Time

Here, the time is the round-trip time for the sound, so the distance to the mosquito is half the total distance:

Total Distance = 340 m/s × 0.0250 s = 8.5 m

Distance to the Mosquito = Total Distance / 2 = 8.5 m / 2 = 4.25 m

Therefore, the mosquito is 4.25 meters away from the bat.

Answer:

Acceleration of gravity on Noveria = 4.4 m/s²

Explanation:

Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.

We have weight, W = mg

Acceleration due to gravity, g = 9.81m/s²    

799 = m x 9.81

Mass of Shepard, m = 81.45 kg            

She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.

We have weight, W = mg'

                 356 = 81.45 xg'

Acceleration of gravity on Noveria, g' = 4.4 m/s²

Answer:

Speed of bullet = 389.61 m/s.

Explanation:

Considering the vertical motion of bullet

Initial vertical speed = 0 m/s

Vertical displacement = 2.9 cm = 0.029 m

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    0.029 = 0 x t + 0.5 x 9.81 x t²

    t = 0.077 s

So ball hits the target after 0.077 s.

Now considering the vertical motion of bullet

Time = 0.077 s

Horizontal displacement = 30 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    30 = u  x 0.077 + 0.5 x 0 x 0.077²

     u = 389.61 m/s

Speed of bullet = 389.61 m/s.

Answer:

390 m/s

Explanation:

Let the horizontal speed be u.

Horizontal distance , x = 30 m

Vertical distance, y = 2.9 cm

Let time taken be t.

Use second equation of motion in vertical direction

H = uy × t + 1/2 gt^2

0.029 = 0.5 × 9.8 × t^2

t = 0.077 s

Horizontal distance = horizontal velocity × time

30 = u × 0.077

u = 390 m/s

Answer:

Part a)

P = 8.23 kg m/s

Part b)

v = 3.05 m/s

Explanation:

Part a)

momentum of cart 1 is given as

[tex]P_1 = m_1v_1[/tex]

[tex]P_1 = (2.7)(3.7) = 9.99 kg m/s[/tex]

Momentum of cart 2 is given as

[tex]P_2 = m_2v_2[/tex]

[tex]P_2 = (1.1)(-1.6) = -1.76 kg m/s[/tex]

Now total momentum of both carts is given as

[tex]P = P_1 + P_2[/tex]

[tex]P = 8.23 kg m/s[/tex]

Part b)

Since two carts are moving towards each other due to mutual attraction force and there is no external force on two carts so here momentum is always conserved

so here we will have

[tex]P_i = P_f[/tex]

[tex](2.7 kg)v = 8.23[/tex]

[tex]v = 3.05 m/s[/tex]

Answer:

[tex]6.35\cdot 10^{-9} W[/tex]

Explanation:

The relationship between power and intensity of a sound is given by:

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

where

I is the intensity

P is the power

A is the area considered

In this problem, we know

[tex]A=2.90\cdot 10^{-3}m^2[/tex] is the surface area of the ear

[tex]I = 2.19\cdot 10^{-6} W/m^2[/tex] is the intensity of the sound

Re-arranging the equation, we can find the power intercepted by the ear:

[tex]P=IA=(2.19\cdot 10^{-6} W/m^2)(2.90\cdot 10^{-3} m^2)=6.35\cdot 10^{-9} W[/tex]

The equation of motion of a pendulum is:

[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,[/tex]

where [tex]\ell[/tex] it its length and [tex]g[/tex] is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for small angles ([tex]\theta \ll 1[/tex]), we can use:

[tex]\sin\theta \simeq \theta.[/tex]

Additionally, let us define:

[tex]\omega^2\equiv\dfrac{g}{\ell}.[/tex]

We can now write:

[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.[/tex]

The solution to this differential equation is:

[tex]\theta(t) = A\sin(\omega t + \phi),[/tex]

where [tex]A[/tex] and [tex]\phi[/tex] are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

[tex]T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.[/tex]

This justifies that the period depends only on the pendulum's length.

Answer:

0.84 Ω

Explanation:

r = mean radius of the turn = 6.5 mm

n = number of turns of copper wire = 150

Total length of wire containing all the turns is given as

L = 2πnr

L =  2 (3.14)(150) (6.5)

L = 6123 mm

L = 6.123 m

d = diameter of the wire = 0.4 mm = 0.4 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.4 x 10⁻³)²

A = 1.256 x 10⁻⁷ m²

ρ = resistivity of copper = 1.72 x 10⁻⁸ Ω-m

Resistance of the coil is given as

[tex]R = \frac{\rho L}{A}[/tex]

[tex]R = \frac{(1.72\times 10^{-8}) (6.123))}{(1.256\times 10^{-7}))}[/tex]

R = 0.84 Ω

Answer:

C

Explanation:

Givens

m = 2 kg

F = 2 * 9.81

F =  19.62 N

x = 0.15 m

Formula

F = k*x

Solution

19.62 = k*0.15

k = 19.62/0.15

k = 130.8 which rounded to the nearest given answer is C