A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS, with respect to the water. The swimmer travels a distance D in a time tOut. The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS, with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn. What is tOut in terms of vR, vS, and D, as needed?

Answer:

General Expression: E = kql/(l² + r²)^(3/2)

(a) 6.3 MN/C

(b) 22.8 MN/C

(c) 6.1 MN/C

(d) 0.63 MN/C

Explanation:

The general expression for electric field along axis of a uniformly charged ring is:

E = kqL/(L² + r²)^(3/2)

where,

E = Electric Field Strength = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q = Total Charge = 71 μC = 71 x 10⁻⁶ C

L = Distance from center on axis

r = radius of ring = 10 cm = 0.1 m

(a)

L = 1 cm = 0.01 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)

E = (6390 N.m³/C)/(0.00101 m³)

E =  6.3 x 10⁶ N/C = 6.3 MN/C

(b)

L = 5 cm = 0.05 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)

E = (31950 N.m³/C)/(0.00139 m³)

E =  22.8 x 10⁶ N/C = 27.4 MN/C

(c)

L = 30 cm = 0.3 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)

E = (191700 N.m³/C)/(0.03162 m³)

E =  6.1 x 10⁶ N/C = 6.1 MN/C

(d)

L = 100 cm = 1 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)

E = (639000 N.m³/C)/(1.015 m³)

E =  0.63 x 10⁶ N/C = 0.63 MN/C

Answer:

The description of that same situation has been listed throughout the explanation segment below.

Explanation:

So that the above seems to be the right answer.

Answer:

Explanation:

1.  [tex]V_{x}[/tex] = [tex]V_{0}[/tex] * cos[tex]\alpha[/tex] ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. [tex]V_{y}[/tex] = [tex]V_{0}[/tex] * sin[tex]\alpha[/tex] ⇒ 16* sin32 ≈ 9.4 m/s

3. [tex]y_{max}[/tex] = [tex]\frac{v_{0}^2*sin^2\alpha}{2g}[/tex]= [tex]\frac{16^2*sin^232}{2*9.8}[/tex] (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

[tex]y_{max}[/tex] ≈ 3.6677+1.5 ≈ 5.2m

4.  [tex]x_{max}[/tex] = [tex]\frac{v_{0}^2*sin(2\alpha)}{g}[/tex]=[tex]\frac{16^2*sin(2*32)}{9.8}[/tex] ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline.  Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left.  Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

Answer:

14.68m/s

Explanation:

As per the question, the data provided is as follows

Mass = M = 0.148 kg

Height = h = 11 m

Initial velocity = U = 0 m/s

Final velocity = V

Gravitational force = F

Mass = M

Based on the above information, the speed that hit to the ground is

As we know that

Work to be done = Change in kinetic energy

[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]

[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]

[tex]V^2 - U^2 = 2gh[/tex]

[tex]V^2 - 0 = 2gh[/tex]

[tex]V = \sqrt{2 g h}[/tex]

[tex]= \sqrt{2\times9.8\times11}[/tex]

= 14.68m/s

Answer:

Increases

Explanation:

Due to an increase in temperature, molecules within the medium will vibrate more vigorously, meaning that the rate of chemical reactions generally increases with temperature due to an increase in kinetic energy. Because sound is a form of kinetic energy, it is safe to assume that the speed of sound waves increases with temperature.

Answer:

A- increases because The particles bump into each other more often.

Explanation:

Just took the test

Answer:

(A) object B has the largest mass because it has the least acceleration

(B) object C has the smallest mass because it has the largest acceleration

(C) mass A : mass B = 3 : 5

Explanation:

Given;

acceleration of object A = 5 m/s²

acceleration of object B = 3 m/s²

acceleration of object C = 7 m/s²

A constant force, F

According to Newton's second law of motion;

F = ma

m = F / a

Mass of object A:

m = F / 5

Mass of object B:

m = F / 3

Mass of object  C:

m = F / 7

(A). Which object has the largest mass:

object B has the largest mass because it has the least acceleration

(B). Which object has the smallest mass:

object C has the smallest mass because it has the largest acceleration

(C). What is the ratio of mass A to mass B;

mass A = F / 5

mass B = F / 3

[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]

A. The Object B has largest mass.

B. The Object A has smallest mass.

C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]

The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.

                   [tex]F=ma\\\\m=\frac{F}{a}[/tex]

                     [tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]

Learn more about the acceleration of object here:

https://brainly.com/question/460763

Answer:

The value of m is 2635.294 grams.

Explanation:

Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:

[tex]f = \frac{\omega}{2\pi}[/tex]

Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.

For a mass-spring system under simple harmonic motion, the angular frequency is:

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

Where:

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m[/tex] - Mass, measured in kilograms.

The following equation is obtained after replacing angular frequency in frequency formula:

[tex]f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }[/tex]

As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:

[tex]f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}[/tex]

If [tex]m_{2} = m_{1} + 700\,g[/tex], [tex]f_{1} = 0.72\,hz[/tex] and [tex]f_{2} = 0.64\,hz[/tex], the resulting expression is simplified and then initial mass is found after clearing it:

[tex]f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}[/tex]

[tex]f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)[/tex]

[tex]\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g[/tex]

[tex]\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g[/tex]

[tex]m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}[/tex]

[tex]m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}[/tex]

[tex]m_{1} = 2635.294\,g[/tex]

The value of m is 2635.294 grams.

Answer:

a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]

Where:

[tex]m_{w}[/tex] - Mass of water, measured in kilograms.

[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].

[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].

Then,

[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]

[tex]Q_{needed} = 180748800\,J[/tex]

The energy needed in kilowatt-hours is:

[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]

[tex]Q_{needed} = 50.208\,kWh[/tex]

The electric energy required to heat up the water is:

[tex]E = \frac{50.208\,kWh}{0.75}[/tex]

[tex]E = 66.944\,kWh[/tex]

Lastly, the cost of heating a hot tub is: (USD - US dollars)

[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]

[tex]E = 6.024\,USD[/tex]

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]

[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]

[tex]i = 88.200\,A[/tex]

Answer:

The force is  [tex]F = 2400 \ N[/tex]

Explanation:

The diagram for this question is shown on the first uploaded image

 From the question we are told that

   The mass of the dancer is  [tex]m_d = 60 \ kg[/tex]

From the diagram the

      The first distance is [tex]l_1 = 20 \ cm[/tex]

      The second distance is  [tex]l_2 = 5 \ cm[/tex]

At equilibrium the moment about the center of the dancers  feet  is mathematically represented as

      [tex]F * l_2 - (mg* l_1)[/tex]

   Where [tex]g= 10 \ m/s^2[/tex]

substituting values

      [tex]F * 5 - (60* 9.8 * 20)[/tex]

=>    [tex]F = \frac{60 * 10 * 30}{5}[/tex]

=>      [tex]F = 2400 \ N[/tex]

Answer:

Explanation:

The kinetic energy possessed by the block when sliding will be equal to the energy needed to compress the string.

Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²

m = mass of the block (in kg) = 2.8 kg

v = speed of the block (in m/s) = 2.4 m/s

k = force constant of the spring

e = extension (in metres) = 0.056m

Since KE = energy stored in the spring

1/2 mv² = 1/2 ke²

mv² = ke²

2.8(2.4)²  = k(0.056)²

16.128 = 0.003136k

k = 16.128/0.003136

k =  5,142.86

The force constant of the spring is 5,236.36

The force that constant of the spring is 5,142.86.

The kinetic energy that should be possessed by the block at the time when sliding will be equivalent to the energy required to compress the string.

Here

Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²

m = mass of the block (in kg) = 2.8 kg

v = speed of the block (in m/s) = 2.4 m/s

k = force constant of the spring

e = extension (in metres) = 0.056m

Since KE = energy stored in the spring

So,

1/2 mv² = 1/2 ke²

mv² = ke²

Now

2.8(2.4)²  = k(0.056)²

16.128 = 0.003136k

k = 16.128/0.003136

k =  5,142.86

Learn more about force here: https://brainly.com/question/3398162

Complete Question

An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters )  is the airplane above the ground 13 seconds after takeoff?

Answer:

The height  is [tex]H = 318.5 \ m[/tex]

Explanation:

From the question we are told that

   The speed at which the plane takes off is  [tex]u = 49 \ m/s[/tex]

      The angle at which it takes off is  [tex]\theta = 30 ^o[/tex]

        The time taken is [tex]t = 13 s[/tex]

The vertical distance traveled is  mathematically represented as

          [tex]H = u sin \theta t[/tex]

Substituting values  

         [tex]H = (49) * sin (30) *13[/tex]

        [tex]H = 318.5 \ m[/tex]

Answer :v=480m400s=1.2ms

2002+4802=H2  

The hypotenuse  H=520m  

A quicker way to get the length of the hypotenuse is to recognize that this is a simple 5–12–13 triangle where the sides are multiples of 5, 12, and 13:

5(40) = 200m, 12(40)= 480m, 13(40)= 520m

We know that the swimmer travelled 520 m in 400 seconds, so her average speed was:

VR=520m400sec=   1.3ms

hope i got it right!! xx

Explanation:

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

Answer:

Explanation:

For a convex mirror, the value of its image distance and its focal length are negative.

using the mirror formula 1/f = 1/u+1/v

f is the focal length = Radius of curvature/2 = 0.560/2

f= 0.28m

u is the object distance = 10.6m

v is the position of the image = ?

On substitution;

1/0.28 = 1/10.6 + 1/-v

3.57 = 0.094 - 1/v

3.57 - 0.094 = -1/v

3.476 = -1/v

v = -1/3.476

v = -0.2877m

B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.

C) Magnification = Image distance/object distance

Magnification  = 0.2877/10.6

Magnification = 0.0271

Answer:

The  drag coefficient is  [tex]D_z = 1.30512[/tex]  

Explanation:

From the question we are told that

     The density of air is  [tex]\rho_a = 1.21 \ kg/m^3[/tex]

     The diameter of bottom part is  [tex]d = 0.15 \ m[/tex]

The  power trend-line  equation is mathematically represented as

      [tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]

let assume that the velocity is  20 m/s

Then

      [tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]

       [tex]F_{\alpha } = 5.1453 \ N[/tex]

The drag coefficient is mathematically represented as

      [tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]

Where  

     [tex]F_{\alpha }[/tex] is the drag force

      [tex]\rho[/tex] is the density of the fluid

       [tex]v[/tex] is the flow velocity

       A is the area which mathematically evaluated as

       [tex]A = \pi r^2 = \pi \frac{d^2}{4}[/tex]

substituting values

     [tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]

     [tex]A = 0.0176 \ m^2[/tex]

Then

   [tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]

   [tex]D_z = 1.30512[/tex]  

Answer:

a) The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex], b) The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Explanation:

This question is incomplete and complete version will be presented herein:

A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring (b) Find the angular frequency of the block 's vibrations.

a) Since spring is hung from the ceiling and is stretched by action of gravity on 0.573 kilogram block. According to the Hooke's Law, force experimented by the spring is directly proportional to elongation. An expression describing the phenomenon is presented and described below: (System at equilibrium - Newton's Second Law)

[tex]m\cdot g = k\cdot \Delta x[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]\Delta x[/tex] - Spring linear deformation, measured in meters.

Now, the spring constant is cleared in this equation and outcome is computed: ([tex]m = 0.573\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta x = 0.198\,m[/tex])

[tex]k = \frac{m\cdot g}{\Delta x}[/tex]

[tex]k = \frac{(0.573\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.198\,m}[/tex]

[tex]k = 28.381\,\frac{N}{m}[/tex]

The spring constant of the spring is [tex]28.381\,\frac{N}{m}[/tex].

b) Let suppose that mass-spring system is experimenting a simple harmonic motion, so that angular frequency is equal to:

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

Given that [tex]k = 28.381\,\frac{N}{m}[/tex] and [tex]m = 0.573\,kg[/tex], the angular frequency, measured in radians per second, of the block is:

[tex]\omega = \sqrt{\frac{28.381\,\frac{N}{m} }{0.573\,kg} }[/tex]

[tex]\omega = 7.038\,\frac{rad}{s}[/tex]

The angular frequency of the block is [tex]7.038\,\frac{rad}{s}[/tex].

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.

The given parameters;

The volume of the bobber is calculated as follows;

[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]

The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;

[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]

The volume of the lead is calculated as follows;

[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]

The buoyant force experienced by the lead due to the volume submerged is calculated as follows

[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]

The total buoyant force is calculated as;

[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]

The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;

[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]

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Answer:

correct answer is c

v = -0.8 m / s

Explanation:

This is a problem of quantity of movement, for this we must define a system formed by the two cars, so that the forces during the collision are internal and therefore the quantity of movement is conserved

initial

    p₀ = m₁ v₁ - m₂ v₂

final

    = (m₁ + m₂) v

We have taken the direction to the right as positive

    p₀ =p_{f}

    m₁ v₁ - m₂ v₂ = (m₁ + m₂) v

    v = (m₁ v₁ - m₂ v₂) / (m₁ + m₂)

we calculate

    v = (2  4 - 8  2) / (2 + 8)

    v = (8 -16) / 10

     v = -0.8 m / s

the negative sign indicates that the set is moving to the left

correct answer is c

Answer:  If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.

Answer:

If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?

Charge will oscillate in the tank's capacitor and inductor.

Explanation:

Answer:

49.725× 10^-24J

Explanation:

The Energy associated with a Photon us defined as;

E = hf

Where h is Planck's constant = 6.63× 10^-34m2kg/s

f is the frequency= 7.5 x 10^14 Hz

Hence

E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex]     (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex]    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]

The mass of the astronaut is 48.48 kg

Answer:

(a)  emf = 1.18 mV

(b) counter-clockwise sense

Explanation:

(a) The induced emf is given by the following formula:

[tex]emf=-\frac{d\Phi_B}{dt}[/tex]     (1)

where:

ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)

A = πr^2

B = 0.800 T

You replace the expression for the magnetic flux in the equation (1):

[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]

A1: initial area

A2: final area

t2-t1: time interval  = 9.0s

Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m

[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]

You calculate the initial area A1:

[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]

After 9.0 second the circumference will be:

[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]

the new radius and the final area is:

[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]

[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]

Finally, you replace in the equation (1):

[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]

The induced emf in the circular loop is 1.18mV

(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.

Answer:

A. 2.083 MV/m from anode to cathode.

B. 93648278.15 m/s

C. 2.5x10^-5 C and there are about 1.56x10^14 electrons

D. 4x10^-15 Joules

Explanation:

Voltage V across plate is 25 kV = 25x10^3 V

Distance apart x = 1.2 cm = 1.2x10^-2 m

A. Electric field strength is the potential difference per unit distance

E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m

= 2.083 MV/m

B. Energy of electron is electron charge times the voltage across

i.e eV

Charge on electron = 1.6x10^-19 C

Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules

Mass of electron m is 9.12x10^-31 kg

Kinetic energy of electron = 0.5mv^2

Where v is the speed

4x10^-15 = 0.5 x 9.12x10^-31 x v^2

v^2 = 8.77x10^15

v = 93648278.15 m/s

C. From Q = CV

Q = charge

C = capacitance = 1 nF 1x10^-9 F

V = voltage = 25x10^3 V

Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C

Total number of electrons = Q/e

= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons

D. To push electron from cathode to anode, I'll have to do a work of about

4x10^-15 Joules

Answer: Tech A is correct

Explanation:

Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.

Answer:

a. TRUTH

b. FALSE

c. TRUTH

d. FALSE

Explanation:

The emf (electromagnetic force) is generated in a loop or solenoid by the change in the magnetic flux in a closed conductor path (for example, a wire).

This can be noted in the following formula, which is known as the Lenz's law:

[tex]emf=-N\frac{d\Phi_B}{dt}=-N\frac{d(AB)}{dt}[/tex]   (1)

Then, the change, in time, of the area of the conductor, or the change in the magnitude of the magnetic field, the induced emf acquires different values. Furthermore, the loops have a resistance, then, a current can be measured when an emf is induced.

Based on this information you have:

a. an induced emf is caused by a changing magnetic flux. TRUTH

b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field. FALSE

c. an induced emf can be observed by measuring the current that is created. TRUTH

d. an induced emf and conventional induced current are in opposite directions. TRUTH (the minus sing in the equation (1) )

Answer:

This example shows that alloys are stronger than either of it's parent materials by themselves.

Explanation:

Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.

This is because an alloy is a mixture of two or more elements, where at least one element is a metal.

Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.

Answer:

The nature of the wave formed is a transverse  progressive wave.

Explanation:

A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.

A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.

The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.