The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alternately stretches, as you bring your weight down onto your forward foot, and contracts to push you off the ground. A 70 kg runner has an Achilles tendon that is 15 cm long and has a cross-section area of 110 mm² typical values for a person of this size. 1. By how much will the runner's Achilles tendon stretch if the force on it is 8.0 times his weight? Young's modulus for tendor is 0.15 x 10¹⁰N/m². Express your answer to two significant figures and include the appropriate units. 2. What fraction of the tendon's length does this correspond.

Answer:

Speed by walking is 33 miles per hour

And speed of by car is 88 miles per hour

Explanation:

We have given that it takes same time to walk 1212 miles as 3232 miles by car

Now let the speed by walk is x

As speed by car is 55 miles per hour faster than by walk = x+55

As time is same and we know that time is given as [tex]time=\frac{distance}{speed}[/tex]

So [tex]\frac{1212}{x}=\frac{3232}{x+55}[/tex]

[tex]1212(x+55)=3232x[/tex]

[tex]1212x+66660=3232x[/tex]

x = 33 miles per hour

So speed by walking is 33 miles per hour

And so speed of car = 33+55 =88 miles per hour

Answer:

[tex]\frac{v_p}{v_a}=1.268[/tex]

Explanation:

1) Basic concepts

Apogee: is the maximum distance for an object orbiting the Earth. For this case the distance for the apogee would be the sum of radius for the earth and 2289 km.

Perigee: Minimum distance of an object orbiting the Earth. For this case the distance for the perigee would be the sum of radius for the earth and 459 km.

A good approximation for these terms are related to the figure atached.

Isolated system: That happens when the system is not subdued to an external torque, on this case the change of angular momentum would be 0.

2) Notation and data

[tex]r_{earth}[/tex] radius for the Earth, looking for this value on a book we got [tex]r_{earth}=6.37x10^{3}km[/tex]

[tex]r_p =459km +r_{earth}=459km +6.37x10^3 km=6.829x10^3 km[/tex], represent the radius for perigee

[tex]r_a =2289km +r_{earth}=2289km +6.37x10^3 km=8.659x10^3 km[/tex], represent the radius for perigee

[tex]T=112.7min[/tex], period

[tex]v_p[/tex] velocity of the perigee

[tex]v_p[/tex] velocity of the apogee

[tex]r=\frac{v_p}{v_a}[/tex] is the variable of interest represented the ratio for the speed of the perigee to the speed of the apogee

3) Formulas to use

We don't have torque from the gravitational force since is a centered force. So the change of momentum is 0

[tex]\Delta L=0[/tex]   (1)

[tex]L_a=L_p[/tex]   (2)

From the definition of angular momentum we have [tex]L_i=mv_i r_i[/tex], replacing this into equation (2) we got:

[tex]mv_a r_a =mv_p r_p[/tex]   (3)

We can cancel the mass on both sides

[tex]v_a r_a =v_p r_p[/tex]   (4)

And solving [tex]\frac{v_p}{v_a}[/tex] we got:

[tex]\frac{v_p}{v_a}=\frac{r_a}{r_p}[/tex]   (5)

And replacing the values obtained into equation (5) we got:

[tex]\frac{v_p}{v_a}=\frac{8.659x10^3 km}{6.829x10^3 km}=1.268[/tex]  

And this would be the final answer [tex]\frac{v_p}{v_a}=1.268[/tex]

Answer:

Difference threshold or also Just Noticeable Difference

Explanation:

The above mentioned case between room mates, where one room mate was able to detect a minute change in volume shows an instance of the difference threshold.

Difference threshold can be defined as stimulation at its minimum level that can be detected by an individual almost 50 % of the times.

It is the lowest possible level of sound that is detectable by a person.

That is what happened in the mentioned case that when the volume was increased from 14 to 15, Amber was able to detect it.

Answer:

Van der waals forces.

Explanation:

When we spread glue to stick any two substances as A and B with adhesives C. then there are adhesive force between substance A and C and adhesive force between substance B and C and cohesive force between C itself will act. In all adhesive and cohesive forces van der waals forces will apply at molecular level because there is no chemical bonding between adhesive and surface but lots of small attractive forces.

Answer:

Van der Waals force

Explanation:

The technical name given to the force that holds glue to its bonding materials is called Van der Waals force.

The forces of Van der Waals is defined by  attraction and repulsion between atoms, molecules, and surfaces and other intermolecular forces. They differ from covalent and ionic bond in that they are caused by correlations in the varying polarizations of the nearby particles (as a result of quantum dynamics).

Answer:45.24 m/s

Explanation:

Given

Height of Building h=24 m

Range of ball R=100 m

Considering Vertical motion of ball

using [tex]y=u_yt+\frac{a_yt^2}{2} [/tex]

initial vertical velocity is zero therefore [tex]u_y=0[/tex]

[tex]24=0+\frac{9.8\times t^2}{2}[/tex]

[tex]t=\sqrt{\frac{48}{9.8}}[/tex]

[tex]t=2.21 s[/tex]

Now considering Horizontal Motion

[tex]R=u_xt+\frac{a_xt^2}{2}[/tex]

[tex]100=u_x\times 2.21+0[/tex]  , as there is no horizontal acceleration

[tex]u_x=45.24 m/s[/tex]

The physics problem can be addressed by using the principles of projectile motion. The time of flight determined by the vertical motion is used to calculate the horizontal initial velocity. The initial velocity of the ball is approximately 45.24m/s.

This is a problem in Physics based on the principles of Projectile Motion. We need to determine the initial velocity of the ball. The key point in this problem is that the horizontal motion of the projectile (in this case, the ball) is determined purely by the initial horizontal velocity, and is unaffected by the vertical motion. This is called the independence of the horizontal and vertical motions.

The time the ball is in the air is governed entirely by its vertical motion. Thus, we can find the time of flight by using the equation for vertical motion: y = 1/2gt², where y is the vertical displacement (24m in this case), g is the acceleration due to gravity (9.81 m/s²), and t is the time. So, t = sqrt(2y/g) = sqrt(2*24/9.81) = 2.21s.

Using this time, we can find the initial horizontal velocity using the equation for horizontal motion: x = vxt where x is the horizontal displacement (100m in this case), vx is the horizontal velocity, and t is the time. Rearranging the equation we get: vx = x/t which is approximately 45.24m/s . So, the initial velocity of the ball is around 45.24m/s

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Answer:

c. Jupiter's greater mass compresses it more, thus increasing its density.

Explanation:

The mass of Jupiter is greater in its interior, this mass compresses Jupiter to some extent. Thus, its density is increased. Now, more mass is compressed in the lesser volume. Hence, its size does not increase enormously. On the other hand the mass of Saturn is lesser  and also density lower. this gives Saturn a reasonably higher volume.

Hence, option C is correct.

Saturn appears almost as large as Jupiter due to its lower mass leading to less gravitational compression of its hydrogen and helium composition, resulting in a larger volume for its mass. The correct answer from the provided choices is c. Jupiter's greater mass compresses it more, thus increasing its density.

The reason why Saturn appears almost as large as Jupiter despite having a smaller mass is largely due to its composition and structure. We see that Jupiter, which is 318 times more massive than Earth, has a significantly higher density because its greater mass compresses the planets' internal hydrogen and helium more than Saturn, which is about 25% as massive as Jupiter. In comparison, Saturn's lower mass results in less compression and hence a larger volume for its mass. So, the correct answer is c. Jupiter's greater mass compresses it more, thus increasing its density.

Jupiter and Saturn are both composed mainly of hydrogen and helium, which become liquid at great depths due to their large size and the resulting high pressure. However, Saturn, with its lower density, is the least dense planet in the solar system, even less dense than water. Contrary to the choices provided, Saturn's rings do not play a significant role in making the planet appear larger in terms of its physical size, and the effect of distance from the Sun on the compactness of the planets is negligible compared to the impact of their massive gravitational compression.

Answer:

h=0.425 L

Explanation:

Given that

θ =  54°

Coefficient of friction μ = 0.32

Mass of rod = m

Lets take mass of man = M  = 4 m

C is the center of mass of the rod.

By balancing force in y and x direction

R= Fr

R = Fr=  μ N

N = mg + Mg = mg + 4 m g                     ( M =4m)

N = 5 m g

Lets take distance cover by man is h along rod before sliding

Now taking moment about the lower end

M g  h cosθ + m g cosθ L/2  =  R L sinθ

2 M g  h cosθ + m g cosθ L  = 2 R L sinθ

Now by putting the value of R  and M

8 m g  h cosθ + m g cosθ L  = 2  μ N L sinθ

8 m g  h cosθ + m g cosθ L  = 10 m g μ  L sinθ

8   h cosθ +  cosθ L  = 10  μ  L sinθ

8 h + L = 10 μ L tanθ

Now putting the value of θ  and μ

8 h + L = 10 x 0.32 x tan54° x L

8 h + L = 4.4 L

8 h = 3.4 L

h=0.425 L

Final answer:

The work done by gravity on a block dropped onto a spring is 0.22638 J, the spring does 1.815 J of work compressing, and if the speed at impact is doubled, the maximum spring compression becomes 22 cm. The speed of the block before impact cannot be determined without the drop height.

Explanation:

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.0 N/cm. The block becomes attached to the spring and compresses it 11 cm before momentarily stopping.

(a) To calculate work done by gravity, use Work = mgh, where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s²), and h is the height (11 cm = 0.11 m, since we need consistent units). The work done by gravity is (0.21 kg)(9.8 m/s²)(0.11 m) = 0.22638 J (since potential energy lost by the block is equal to the work done by gravity).

(b) The work done by the spring is Work = 1/2 kx², converting k to N/m gives us 300 N/m. The compression x is 0.11 m, so the work done is (1/2)(300 N/m)(0.11 m)² = 1.815 J.

(c) Since we're assuming friction is negligible and using energy conservation, the potential energy (mgh) at the beginning will be equal to the kinetic energy (1/2 mv²) just before impact. Solving for v gives v = √(2gh), where h is the drop height. However, without the drop height, we cannot calculate the exact velocity.

(d) Doubling the speed will increase the kinetic energy by a factor of four (since KE = 1/2 mv²). To find the new compression distance, we set the new kinetic energy equal to the spring potential energy (1/2 kx²) and solve for x. The maximum compression x will be twice the original compression, or 22 cm.

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

[tex]c(50 + 23c) = 10.32[/tex]

[tex]23c^2 + 50c - 10.32 = 0 [/tex]

[tex]c \approx 0.19m[/tex]

Answer:

W = 112.76MJ

Explanation:

the work is:

[tex]W = F_xD[/tex]

where [tex]F_x[/tex] is the force executed in the direction of the displacement and the d the displacement.

so:

W = 12000Ncos(20)(10000)

we use the cos of the angule because it give us the proyection in the axis x of the force, that means the force in the direction of the displacement.

W = 112.76MJ

Answer:

183.75641 Joules

Explanation:

F = Force of the vacuum cleaner = 84.5 N

s = Displacement of the vacuum cleaner = 2.62 m

[tex]\theta[/tex] = Angle the force makes with the horizontal = 33.9°

Work done is given by

[tex]W=F\times scos\theta\\\Rightarrow W=84.5\times 2.62\times cos33.9\\\Rightarrow W=183.75641\ J[/tex]

The work done by the force of the vacuum cleaner is 183.75641 Joules

The work done by the man pushing the vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m is approximately 184.8 joules.

To calculate the work done by a force, you can use the formula W = F × d × cos(θ), where W is the work done, F is the magnitude of the force, d is the distance the object moves, and θ is the angle the force makes with the horizontal direction of movement. In this case, the man pushes a vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m. We multiply the force by the distance and the cosine of the angle to find:

W = 84.5 N × 2.62 m × cos(33.9°)

Calculating cosine of 33.9 degrees and multiplying with the force and distance, we get:

W = 84.5 × 2.62 × 0.8326 ≈(joules)

The man does approximately 184.8 joules of work pushing the vacuum cleaner.

Answer:

Part a)

[tex]F_f = 692 N[/tex]

Part b)

[tex]F_f = 932 N[/tex]

Explanation:

Part A)

As we know by Force equation on the losing player

[tex]F - F_f = ma[/tex]

so we will have

[tex]800 - F_f = 90\times 1.20[/tex]

[tex]800 - F_f = 108[/tex]

[tex]F_f = 692 N[/tex]

Part b)

As we know that the winning player is also moving with same acceleration

so we will have

[tex]F_f - F = ma[/tex]

[tex]F_f - 800 = 110\times 1.20[/tex]

[tex]F_f = 932 N[/tex]

Answer:

The measure of angle B equals the measure of angle C.

Explanation:

When a ray of light falls on a smooth polished surface and the light ray bounces back, it is called the reflection of light.

There are two laws of reflection;

1) The incident ray and reflected ray and normal all lie in the same plane.

2) The angle of incidence is equal to angle of reflection.

Here angle of incidence is represented by B and angle of reflection is by C.

So according to law of reflection the measure of angle B equals the measure of angle C.

Answer:

The measure of angle B equals the measure of angle C

Explanation:

"B" is the correct answer on edge

Answer:168 W

Explanation:

Given

Power needed [tex]P=42 W[/tex]

initial Launch velocity is v

Energy of projectile when it is launched [tex]E=\frac{1}{2}mv^2[/tex]

[tex]Power=\frac{Energy}{time}[/tex]

[tex]Power=\frac{E}{t}[/tex]

[tex]42=\frac{\frac{1}{2}mv^2}{t}--------1[/tex]

Power when it is launched with 2 v

[tex]E_2=\frac{1}{2}m(2v)^2=\frac{4}{2}mv^2[/tex]

[tex]P=\frac{2mv^2}{t}---------2[/tex]

Divide 1 & 2 we get

[tex]\frac{42}{P}=\frac{1}{2\times 2}[/tex]

[tex]P=42\times 4=168 W[/tex]    

To accelerate the projectile to twice its launch speed, four times the power is needed.

To find the power needed to accelerate the projectile from rest to a launch speed of 2v in a time of t, we need to recognize that power is directly proportional to the change in kinetic energy. The change in kinetic energy from rest to launch speed v is given by KE = (1/2)mv^2, and the change in kinetic energy from rest to launch speed 2v is given by KE' = (1/2)m(2v)^2 = 4(1/2)mv^2 = 4KE.

Since power is directly proportional to the change in kinetic energy, the power needed to accelerate the projectile to a launch speed of 2v is four times the power needed to accelerate it to a launch speed of v. Therefore, the power needed is 4(42.0 W) = 168.0 W.

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Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus,  atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

The atomic radius of main-group elements decreases as you move to the right across a period due to increased positive charge, while the atomic radius of transition elements remains relatively constant.

The atomic radius of main-group elements decreases as you move to the right across a period because the number of protons in the nucleus increases. This increased positive charge pulls the electrons closer to the nucleus, reducing the size of the atom. In contrast, the atomic radius of transition elements remains relatively constant as you move across a period because their outermost electrons are in different energy levels or subshells. The addition of protons does not significantly affect the size of the atom.

Moving across a period, the number of protons in the nucleus increases, leading to a greater positive charge. This increased positive charge exerts a stronger pull on the electrons, pulling them closer to the nucleus and resulting in a smaller atomic radius.

However, electron shielding, or the repulsion between electrons in different energy levels, also plays a role. As you move across a period, the number of electrons in the same energy level (shell) remains constant, providing consistent shielding effects. This partial counteraction to the increased positive charge contributes to the overall trend of decreasing atomic radius.

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Answer:

P=144.74W

Explanation:

We can model the power output of a resistance by using the following formula:

[tex]P=\frac{V^{2}}{R}[/tex]

Wehre P is the power output, V is the rms voltage and R is the resistance. The resistance of the space heater will remain the same, so we can calculate it from the power output in Germany and its rms voltage. So when solving for R, we get:

[tex]R=\frac{V^{2}}{P}[/tex]

and we can now use the provided data:

[tex]R=\frac{(230V)^{2}}{531.7W}[/tex]

which yields:

R= 99.49 Ω

Once we know what the heater's resistance is, we can now go ahead and calculate the power outpor of the heater in the U.S.

[tex]P=\frac{V^{2}}{R}[/tex]

so

[tex]P=\frac{(120V)^{2}}{99.49\Omega}[/tex]

P=144.74W

Answer:

[tex]\Delta U = 64218.9 J[/tex]

Explanation:

As we know that power of the heater is given as

P = 100 W

now the energy given by the heater for 11 min of time

[tex]E = P \times t[/tex]

[tex]E = 100 \times 11 \times 60[/tex]

[tex]E = 100\times 11 \times 60[/tex]

[tex]E = 66000 J[/tex]

now from 1st law of thermodynamics we know that

[tex]E = \Delta U + W[/tex]

work done under constant pressure condition we have

[tex]W = P \Delta V[/tex]

[tex]W = (3.527 \times 1.01 \times 10^5)(6 - 1) \times 10^{-3}[/tex]

[tex]W = 1781.13 J[/tex]

now from first equation we have

[tex]66000 = 1781.13 + \Delta U[/tex]

[tex]\Delta U = 64218.9 J[/tex]

The change in the internal energy of the system is 67.787 kJ.

The given parameters;

The heat added to the system by the heater;

Q = Pt = 100 x 660 = 66,000 J = 66 kJ

The work done on the system is calculated as follows;

W = PΔV

W = 3.527(6 - 1)

W = 17.64 L.atm

1 Latm = 0.1013 kJ

17.64 Latm = 1.787 kJ

The change in the internal energy of the system is calculated by applying the first law of thermodynamics as follows;

ΔU = Q + W

ΔU = 66 kJ  +  1.787 kJ

ΔU = 67.787 kJ.

Thus, the change in the internal energy of the system is 67.787 kJ.

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Answer:

rise in temperature = 333.53°C

Explanation:

mass of lead bullet, m = 5.9 g

initial velocity, u = 410 m/s

specific heat of lead = 0.126 J/gm - K = 126 J/kg K

Initial kinetic energy of the bullet

K = 1/2 mu^2 = 0.5 x 5.9 x 10^-3 x 410 x 410 = 495.895 J

Half of the kinetic energy is used to raise the temperature of lead bullet

K / 2 = mass of bullet x specific heat of lead x rise in temperature

247.95 = 5.9 x 10^-3 x 126 x rise in temperature

rise in temperature = 333.53°C

Thus, the rise in temperature of lead bullet is 333.53°C.

The final velocity of the truck is 6.33 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: the total momentum of the truck-car system must be conserved before and after the collision (if there are no external forces), so we can write

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]

where:

[tex]m_1 = 3000 kg[/tex] is the mass of the truck

[tex]u_1 = 8 m/s[/tex] is the initial velocity of the truck

[tex]v_1[/tex] is the final velocity of the truck

[tex]m_2 = 500 kg[/tex] is the mass of the car

[tex]u_2 = 0[/tex] is the initial velocity of the car

[tex]v_2 = 10 m/s[/tex] is the final velocity of the car

And by solving the equation for [tex]v_1[/tex], we find the velocity of the truck after the collision:

[tex]v_1 = \frac{m_1 u_1-m_2 v_2}{m_1}=\frac{(3000)(8)-(500)(10)}{3000}=6.33 m/s[/tex]

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The angular velocity of the disk must be 2.25 rpm

Explanation:

The centripetal acceleration of an object in circular motion is given by

[tex]a=\omega^2 r[/tex]

where

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

r is the distance of the object from the axis of rotation

For the space station in this problem, we have

[tex]a=\frac{g}{2}=\frac{9.8}{2}=4.9 m/s^2[/tex] is the centripetal acceleration

The diameter of the disk is

d = 175 m

So the radius is

[tex]r=\frac{175}{2}=87.5 m[/tex]

So, a point on the rim has a distance of 87.5 m from the axis of rotation. Therefore, we can re-arrange the previous equation to find the angular velocity:

[tex]\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{4.9}{87.5}}=0.237 rad/s[/tex]

And this is the angular velocity of any point along the disk. Converting into rpm,

[tex]\omega=0.236 \frac{rad}{s}\cdot \frac{60 s/min}{2\pi rad/rev}=2.25 rpm[/tex]

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When the pulling force points at an angle θ above the horizontal, the frictional force acting on the box is -20N. By using the equation for frictional force and the weight of the box, we can determine that the coefficient of friction is 0.133. To find the angle θ, we use trigonometric ratios and find that it is 150°.

Given that the weight of the box is 150N and the pulling force has a magnitude of 110N, we can determine the angle θ at which the pulling force is directed. Let's assume the angle θ is above the horizontal. The weight of the box, 150N, is equal to the normal force acting on the box. The frictional force between the box and the floor can be calculated as the difference between the force of the pulling and the weight of the box, which is 110N - 150N = -40N. Since the kinetic frictional force acting on the box is twice as large when the pulling force points horizontally, the frictional force when the pulling force points at an angle θ is -20N.

We can use the equation for frictional force, which is F_friction = μN, where F_friction is the frictional force, μ is the coefficient of friction, and N is the normal force. As the frictional force is -20N, we can substitute this value into the equation and solve for the coefficient of friction. Therefore, -20N = μ(150N), which gives us μ = -20N/150N = -0.133. Since the coefficient of friction is always positive, the actual value of μ is 0.133.

Now, let's use trigonometric ratios to find the angle θ. Since the weight of the box acts vertically downward and the pulling force has a horizontal component of 110N and a vertical component of -150N × sin(θ), the vertical components of the weight and the pulling force must cancel each other. Therefore, -150N × sin(θ) = 150N, which simplifies to sin(θ) = -1/2. Taking the inverse sine of -1/2, we get θ = -30° or 150°. However, since the pulling force is directed above the horizontal, the angle must be 150°.

Answer:

The vector a, the answer is the point D

Explanation:

We need to make the make the subtraction from the x,y points. It will be the points of the head minus the points of the tail.

x=(-2 - 0) and y = (3-0) vector = -2 + 3

The Answer is—————————D. a

Answer:

Doug speed will be 65 km/hr

Explanation:

Let the Thor's speed is x km/hr

So Doug's speed = x+1 km/hr

We have given that Doug and Thor take same time to cover 390 km and 384 km respectively

We know that time is given by

[tex]time=\frac{distance}{speed}[/tex]

So time taken by Doug to cover the distance

[tex]time=\frac{390}{x+1}[/tex]

And time taken by Thor to cover the distance

[tex]time=\frac{384}{x}[/tex]

As both times are equal

So [tex]\frac{390}{x+1}=\frac{384}{x}[/tex]

[tex]6x=384[/tex]

[tex]x=64km/hr[/tex]

So Doug speed will be 64+1 = 65 km/hr

Answer:

There is a constant emf induced in the loop.

Explanation:

In the uniform magnetic field suppose the rectangular wire loop of length L and width b is moved out with a uniform velocity v. suppose any instance x length of the loop is out of the magnetic field and L-x length is inside the loop.

Area of loop outside the field = b(L-x)

we know that flux φ= BA

B= magnitude of magnetic field , A=  area

and emf [tex]\epsilon= \frac{d\phi}{dt}[/tex]

[tex]\epsilon=B\frac{dA}{dt}[/tex]

[tex]\epsilon=B\frac{db(L-x)}{dt}[/tex]

[tex]\epsilon=Bb\frac{d(L-x)}{dt}[/tex]

B,b and L are constant and dx/dt = v

⇒ε = -Bbv

which is a constant hence There is a constant emf induced in the loop.

Answer:

v1 = 2.76 m/s and v2 = - 0.32 m/s

Explanation:

m1 = 0.140 kg

m2 = 0.299 kg

u1 = 0.80 m/s

u2 = - 2.28 m/s

Let the speed after collision is v1 and v2.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2

0.112 - 0.68 = 0.14 v1 + 0.299 v2

0.14 v1 + 0.299 v2 = - 0.568 ..... (1)

By the use of coefficient of restitution, the value of e = 1 for elastic collision

[tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]

u2 - u1 = v1 - v2

- 2.28 - 0.8 = v1 - v2

v1 - v2 = 3.08

v1 = 3.08 + v2

Put in equation (1)

0.14 (3.08 + v2) + 0.299 v2 = - 0.568

0.43 + 0.44 v2 = - 0.568

v2 = - 0.32 m/s

and

v1 = 3.08 - 0.32 = 2.76 m/s

Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s

Answer:

7,157

Explanation:

Hi!

Use this equation to solve this problem:

Fb=ρfl*g*Vfl

where

Fb=buoyant force

What do we know?

The known values are highlighted in.

An iceberg (density=917 kg/[tex]m^3[/tex]) is floating in seawater, the volume of the iceberg below the waterline is approximately 7168 [tex]m^3[/tex].

To calculate the volume of the iceberg below the waterline, we must apply the concept of buoyancy. The buoyant force acting on an item that floats in a fluid is equal to the weight of the fluid displaced by the object.

The buoyant force (F_b) may be computed using the formula:

F_b = ρ_fluid * V_submerged * g

W_iceberg = ρ_iceberg * V_iceberg * g

F_b = W_iceberg

ρ_fluid * V_submerged * g = ρ_iceberg * V_iceberg * g

Now,

V_submerged = (917 / 1025 ) * 8000

V_submerged = 0.896 * 8000 [tex]m^3[/tex]

V_submerged ≈ 7168 [tex]m^3[/tex]

Therefore, the volume of the iceberg below the waterline is approximately 7168  [tex]m^3[/tex].

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Answer:

ΔU= *-26 KJ

Explanation:

Given that

Work done by motor W= 30 KJ

Heat gains by motor Q= 4 KJ

Sign convention:

 If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.

If work done by the system then it is taken as positive and if work is done on the system then it is taken as negative.

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in internal energy

Q=Heat transfer

W=Work

Now by putting the values

4 = 30 + ΔU

ΔU= -26 KJ

Answer:

Internal energy ∆U=-26KJ

Explanation:

Given that:

Work done by the motor=+30KJ

Heat gained by the motor=+4KJ

In solving thermodynamical questions it is reasonable to use the sign convention this

Heat is positive if it is added to a system,but becomes negative if the system rejects heats.

Work is positive if the system does work,but becomes negative if work is done on the system.

Using the thermodynamics first law

∆U=Q-W

∆U= 4-30=-26KJ

Answer:

9.1°

Explanation:

Torque = distance from pivot * perpendicular force

15 Nm = 0.5 m * perpendicular force

perpendicular force  = 30 N

So the bertical component of the applied force that caused a turning  effect (torque) was 30 N. Now we use this information to find the angle that would produce a vertical component of  30N from an applied force of 95 N.

Fy = FSinθ

15 = 95 Sinθ

θ = 9.1°

Answer:

The temperature required is near about 3 million kelvin

Explanation:

The brilliance of the star results from the nuclear reaction that take place in the core of the star and radiate a huge amount of thermal energy resulting from the fusion of hydrogen into helium.

For this reaction to take place, the temperature of the star's core must be near about 3 million kelvin.

The hydrogen atoms collide and starts and the energy from the collision results in the heating of the gas cloud. As the temperature comes to near about [tex]1.5\times 10^{7 {\circ}C[/tex], the nuclear fusion reaction takes place in the core of the gas cloud.

The huge amount of thermal energy from the nuclear reaction gives the gas cloud a brilliance resulting in a protostar.

The core temperature necessary for hydrogen fusion to begin in a star is approximately 15,000,000 Kelvin, marking its entry into the main sequence. With hydrogen's exhaustion, fusion of helium and other more complex elements happen, each needing significantly higher temperatures.

Hydrogen fusion, also known as thermonuclear fusion, initiates within a star when its core temperature reaches approximately 15,000,000 Kelvin (K). At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. This process of hydrogen fusion occurring in the core of a star is critical for a star's energy balance and longevity.

A star reaches the main sequence when its core temperature is high enough (about 12 million K) to fuse hydrogen into helium. However, once the hydrogen fuel is exhausted in a star's core, fusion of helium, and later, other more complex elements can occur, with each requiring significantly higher temperatures than previous fusion reactions.

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Answer:[tex]\mu [/tex]=0.48

Explanation:

Given

inclination [tex]\theta =31^{\circ}[/tex]

Acceleration of object[tex]=0.99 m/s^2[/tex]

Now using FBD

[tex]mg\sin \theta -f_r=ma[/tex]

[tex]mg\sin \theta -\mu mg\cos \theta =ma[/tex]

[tex]a=g\sin \theta -\mu g\cos \theta [/tex]

[tex]0.99=5.04-\mu 8.4[/tex]

[tex]\mu 8.4=4.057[/tex]

[tex]\mu =0.48[/tex]