Sprinting up a short flight of stairs is often used as a simple test to compare power output among different athletes. Subject 1, a football player with a body mass of 130 kg, sprints up a flight of stairs and raises his body mass 4.4 m vertically in a time of 3.4 s. Subject 2, a cyclist with a body mass of 75 kg, sprints up the same flight of stairs in 2.8 s. Determine the mechanical work and power for both of these athletes. Which athlete is more powerful in an absolute sense (W)? What if power is expressed relative to body mass (W/kg)? Assuming that the body is 25% efficient, determine the amount of metabolic energy spent during this stair sprint for both athletes (total Kcals, not Kcal/hour; Hint: 1 J = 0.000239 Kcals).

Answer:

E. 5

Explanation:

N₀ = initial total number of radioactive elements number

N = Number of atoms of radioactive element after "n" half lives = N₀ /32

n = number of half lives

Number of atoms of radioactive element after "n" half lives is given as

[tex]N = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]

inserting the values

[tex]\frac{N_{0}}{32} = N_{o}\left ( \frac{1}{2} \right )^{n}[/tex]

[tex]\frac{1}{32} = \left ( \frac{1}{2} \right )^{n}[/tex]

n = 5

Answer:

4 m/s

0.82 s

Explanation:

h = height to which the fox jumps = 81 cm = 0.81 m

v₀ = speed at which the fox leaves the snow

v = speed of the fox at highest point = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

Using the kinematics equation

v² = v₀² + 2 a h

0² = v₀² + 2 (- 9.8) (0.81)

v₀ = 4 m/s

t = amount of time in air while going up

Using the equation

v = v₀ + a t

0 = 4 + (- 9.8) t

t = 0.41 s

T = Total time

Total time is given as

T = 2 t

T = 2 (0.41)

T  = 0.82 s

The speed at which the fox leaves the snow is approximately 3.987 m/s. The fox is in the air for approximately 0.407 seconds.

To calculate the speed at which the fox leaves the snow, we can use the concept of vertical motion and the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Since the fox jumps straight up, the initial velocity is 0 m/s and the displacement is 81 cm (or 0.81 m). Assuming the acceleration due to gravity is 9.8 m/s^2, we can now calculate the final velocity:

v^2 = u^2 + 2as
v^2 = 0^2 + 2(9.8)(0.81)
v^2 = 15.876
v = √15.876
v ≈ 3.987 m/s

The time the fox is in the air can be calculated using the equation v = u + at, where t is the time. Again, the initial velocity is 0 m/s and the acceleration due to gravity is 9.8 m/s^2. Plugging in these values, we have:

v = u + at
3.987 = 0 + (9.8)t
3.987 = 9.8t
t = 3.987/9.8
t ≈ 0.407 s

Answer:

Explanation:

The change in temperature may be calculated from the formula:

Where:

Then, you can make these assumptions or inferences from the conditions stated in the problem:

Then, you can state that, for those samples, ΔT = k / C, i.e. the increase in temperature is inversely related to the specific heat.

That means that the higher the specific heat the lower ΔT, and the lower the specific heat the higher ΔT.

The ranking in decrasing order of specific heat is:

Ranking in increasing order of ΔT:

And since all of them started at the same temperature, that is the ranking in resulting temperature from lowest to highest:

That means that the sample of water, the matter with the highest specific heat capacity (4.19 J/g°C), will reach the lowest temperature, and the sample of iron, the matter with the lowest heat capacity (0.46 J/g°C) will reach the highest temperature.

Answer:

[tex]strain = 1.4 \times 10^{-3} [/tex]

Explanation:

As we know by the formula of elasticity that

[tex]E = \frac{stress}{strain}[/tex]

now we have

[tex]E = 110 GPA[/tex]

[tex]E = 110 \times 10^9 Pa[/tex]

Area = 15.2 mm x 19.1 mm

[tex]A = 290.3 \times 10^{-6}[/tex]

now we also know that force is given as

[tex]F = 44500 N[/tex]

here we have

stress = Force / Area

[tex]stress = \frac{44500}{290.3 \times 10^{-6}}[/tex]

[tex]stress = 1.53 \times 10^8 N/m^2[/tex]

now from above formula we have

[tex]strain = \frac{stress}{E}[/tex]

[tex]strain = \frac{1.53 \times 10^8}{110 \times 10^9}[/tex]

[tex]strain = 1.4 \times 10^{-3} [/tex]

Answer:

a) 0.124 m

b) 0.93 ms⁻¹

c) 0.5 k A² cos ² ( ωt )  

Explanation:

1) Potential energy = U = 0.5 k A² , where A is the amplitude and k = 850 N/m is the spring constant.

0.5 ( 850) (A² ) = 6.5

⇒ A = 0.124 m = Amplitude.

b) From energy conservation,  0.5 m v² =  6.5

⇒ speed = v = 0.93 ms⁻¹

c) If x = A cos ωt ,

Potential energy = 0.5 k A² = 0.5 k A² cos ² ( ωt )  

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]

[tex]\dfrac{d\phi}{dt}[/tex] is the rate of change if magnetic flux.

[tex]\phi=BA\ cos\theta[/tex]

[tex]\theta[/tex] is the angle between the magnetic field and the normal to area vector.

[tex]\theta=90-60=30[/tex]

[tex]\epsilon=NA\dfrac{dB}{dt}\times cos30[/tex]

[tex]\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)[/tex]

[tex]\epsilon=0.0598\ T[/tex]

[tex]\epsilon=59.8\ mT[/tex]

or

EMF = 60 mT

So, the magnitude of  emf induced in the coil is 60 mT. Hence, this is the required solution.

Answer:

Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.

Explanation:

Vertical motion of fish:

 Initial speed, u = 0

 Acceleration, a = 9.81 m/s²

 Displacement, s = 10 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 0² + 2 x 9.81 x 10 = 196.2

    v = 14 m/s

 Final vertical speed = 14 m/s

 Final horizontal speed = initial horizontal speed = 3 m/s

 Final velocity = 3 i - 14 j m/s

 Magnitude

     [tex]v=\sqrt{3^2+(-14)^2}=14.32m/s[/tex]

 Direction

      [tex]\theta =tan^{-1}\left ( \frac{-14}{3}\right )=-77.91^0[/tex]

 Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.

Answer:

N = 768.4 N

Explanation:

As per given FBD we can see that the person inside the elevator have two forces on it

1) Normal force upwards

2) weight downwards

Now from Newton's law of motion we can say

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

[tex]N - mg = ma[/tex]

[tex]N = mg + ma[/tex]

now plug in all values in it

[tex]N = 68(1.5) + 68(9.8)[/tex]

[tex]N = 768.4 N[/tex]

Answer:

Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.

Explanation:

Let line of action be horizontal axis , mass of ball be m and unknown velocity be v.

Here momentum is conserved.

Initial momentum =Final momentum

Initial momentum = m x 6i + m x 0i = 6m i

Final momentum = m x (5.21cos 29.7 i + 5.21sin 29.7 j) + m x v = 4.26 m i + 2.58 m j + m v

4.26 m i + 2.58 m j + m v = 6m i

v = 1.74 i - 2.58 j

Magnitude of velocity [tex]=\sqrt{1.74^2+(-2.58)^2}=3.11m/s[/tex]

Direction,

         [tex]\theta =tan^{-1}\left ( \frac{-2.58}{1.74}\right )=--56^0[/tex]

Velocity is 3.11 m/s at an angle of -56° with respect to the original line of motion.

Answer:

The speed of the water shoot out of the hole is 20 m/s.

(d) is correct option.

Explanation:

Given that,

Height = 20 m

We need to calculate the velocity

Using formula Bernoulli equation

[tex]\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]

Where,

v₁= initial velocity

v₂=final velocity

h₁=total height

h₂=height of the hole from the base

Put the value into the formula

[tex]v_{1}^2=2g(h_{2}-h_{1})[/tex]

[tex]v_{1}=\sqrt{2g(h_{2}-h_{1})}[/tex]

[tex]v_{1}=\sqrt{2\times9.8\times(20-0.005)}[/tex]

[tex]v_{1}=19.7\ m/s= approximate\ 20\ m/s[/tex]

Hence, The speed of the water shoot out of the hole is 20 m/s.

Answer:

Explanation:

The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.

In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.

Here, In the balanced wheat stone bridge

R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.

Resistors in a Wheatstone bridge can be treated as combinations of series and/or parallel resistors for simplification in both balanced and unbalanced bridges. A balanced bridge allows separate treatment of two parallel branches, while unbalanced requires careful analysis. Not all resistor networks can be simplified into series or parallel models.

In a Wheatstone bridge, the resistors can indeed be treated as combinations of series and/or parallel resistors when aiming to simplify calculations or understand the behavior of the circuit. For an unbalanced bridge, resistors are not in simple series or parallel arrangements with respect to the entire circuit due to the bridge not being in equilibrium. However, within certain parts of the bridge, resistors may appear to be in series or parallel with each other. In a balanced bridge, where the bridge is in a state of equilibrium and the central voltmeter reads zero, the two arms of the bridge can be treated separately as two parallel voltage dividers, because no current flows through the meter, effectively decoupling the two parallel branches.

Attempting to simplify a complex resistor network encountered in bridges can indeed be done by identifying and replacing series and parallel resistor combinations step by step until a single resistance value is found. However, this approach cannot always be applied to any arbitrary combination of resistors. Some configurations might contain elements that cannot be reduced to mere parallel or series connections, usually because they form more intricate networks, such as bridges or loops not separable into simpler series or parallel sections.

In conclusion, while more complex connections of resistors in circuits like the Wheatstone bridge can often be broken down into combinations of series and parallel, this is not universally the case for all resistor networks. In certain scenarios, specific techniques or theorems such as Kirchhoff's laws might be required to analyze the circuit effectively.

Answer:

[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Explanation:

As we know that heat flux is given by the formula

[tex]q^n = K\frac{dT}{dx}[/tex]

here we know that

K = thermal conductivity

[tex]\frac{dT}{dx}[/tex] = temperature gradient

now we know that

[tex]q^n = 11 W/m^2[/tex]

also we know that

K = 1.7 W/mK

now we have

[tex]11 = 1.7 \frac{dT}{dx}[/tex]

so temperature gradient is given as

[tex]\frac{dT}{dx} = \frac{11}{1.7} = 6.47 K/m [/tex]

also in other unit it will be same

[tex]\frac{dT}{dx} = 6.47 ^oC/m[/tex]

Also as we can see the equation that heat flux directly depends on the temperature gradient so more is the temperature gradient then more will be the heat flux.

For positive temperature gradient the heat will flow outwards while for negative temperature gradient the heat will flow inwards

Answer:

[tex]1.0672\times 10^{-9}N[/tex]

Explanation:

[tex]G[/tex] = Gravitational constant = 6.67 x 10⁻¹¹

[tex]F[/tex]  = Gravitational force between these spheres

[tex]m_{1}[/tex] = mass of first sphere = 1.60 kg

[tex]m_{2}[/tex]  = mass of second sphere = 16 g = 0.016 kg

[tex]r[/tex]  = distance between the centers of the sphere = 4 cm = 0.04 m

Gravitational force between these spheres is given as

[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]

[tex]F = \frac{(6.67\times 10^{-11})(1.60)(0.016)}{0.04^{2}}[/tex]

[tex]F = 1.0672\times 10^{-9}N[/tex]

Answer:

a) 7.35 x 10¹³ m/s²

b) 5.03 x 10⁻⁸ sec

c) 9.3 cm

d) 6.23 x 10⁻¹⁸ J

Explanation:

E = magnitude of electric field = 418 N/C

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

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

a)

acceleration of the electron is given as

[tex]a = \frac{qE}{m}[/tex]

[tex]a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}[/tex]

a = 7.35 x 10¹³ m/s²

b)

v = final velocity of the electron = 3.70 x 10⁶ m/s

v₀ = initial velocity of the electron = 0 m/s

t = time taken

Using the equation

v = v₀ + at

3.70 x 10⁶ = 0 + (7.35 x 10¹³) t

t = 5.03 x 10⁻⁸ sec

c)

d = distance traveled by the electron

using the equation

d = v₀ t + (0.5) at²

d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²

d = 0.093 m

d = 9.3 cm

d)

Kinetic energy of the electron is given as

KE = (0.5) m v²

KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²

KE = 6.23 x 10⁻¹⁸ J

Answer:

E= 3.4893 N/C

Explanation:

Given s=6.20 m , t=2.50μs, m=9.11*10^-31 Kg  , q= 1.6*10^-19 C

the distance traveled by the electron in time t is

s=ut+0.5at^2

here, u is the initial velocity of the electron, t is time taken and

a is acceleration.

Since the electron is initially at rest u=0

now s=0.5at^2

Therefore a=2s/t^2

also. we know that strength of electric field is

E=ma/q

[tex]E= \frac{2ma}{qt^2}[/tex]

now puting the values we get

[tex]E=\frac{9.11\times 10^-31\times 2\times 6.20}{1.6\times 10^-19\times (4.5\times 10^-6)^2}[/tex]

therefore, E= 3.4865 N/C

The magnitude of the electric field is calculated by first determining the acceleration of the electron and then using the electric force equation to find the electric field. The resulting electric field is 34.8 N/C.

To find the magnitude of the electric field, we first need to calculate the acceleration of the electron. Given that the electron travels a distance of 6.20 m in a time of 4.50 µs (4.50 × 10-6 s), we can use the equations of motion.

Initial velocity, u = 0 (since the electron is released from rest)

Time, t = 4.50 × 10-6 s

Distance, s = 6.20 m

Using the equation of motion: s = ut + 0.5at2

Substitute the values: 6.20 = 0 + 0.5a(4.50 × 10-6)2

6.20 = 0.5a(20.25 × 10-12)

a = 6.20 / (0.5 × 20.25 × 10-12)

a = 6.20 / (10.125 × 10-12)

a = 6.12 × 1011 m/s2

Now, we calculate the electric field using Newton's Second Law, F = ma, and the electric force equation, F = eE, where e is the charge of the electron (1.60 × 10-19 C) and E is the electric field.

ma = eE

(9.11 × 10-31 kg)(6.12 × 1011 m/s2) = (1.60 × 10-19 C)E

(5.57 × 10-19 N) = (1.60 × 10-19 C)E

E = 5.57 × 10-19 N / 1.60 × 10-19 C

E = 3.48 × 101 N/C

Thus, the magnitude of the electric field is 34.8 N/C.

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is [tex]\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}[/tex]

Substituting values

    [tex]3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s[/tex]

Speed of block A after collision = 10 m/s

Option A is the correct answer.

Answer:

8m/s

Explanation:

To find the total mass of chloroform in the second sample, the mass of carbon is scaled proportionally to find the corresponding masses of hydrogen and chlorine. By summing these, the total mass of chloroform is calculated to be 298.525 grams.

To determine the total mass of chloroform in the second sample, we must first understand that chloroform has a known molecular formula of CHCl3. Given that the first sample contains 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen, we can deduce the mass ratios of the elements within chloroform. Using the molecular mass of chloroform, which is 119.37 g/mol, we can calculate the masses of hydrogen and chlorine in the second sample based on the given mass of carbon.

For the second sample: If 12.0 g of carbon is accompanied by 1.01 g of hydrogen and 106.4 g of chlorine, then 30.0 g of carbon should be accompanied by:

Thus, the total mass of chloroform in the second sample would be the sum of the masses of carbon, hydrogen, and chlorine: 30.0 g C + 2.525 g H + 266.0 g Cl = 298.525 g of chloroform.

Parts made from blow molding are plastic, hollow, and thin-walled, such as bottles and containers that are available in a variety of shapes and sizes. Small products may include bottles for water, liquid soap, shampoo, motor oil, and milk, while larger containers include plastic drums, tubs, and storage tanks.

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed [tex]\omega[/tex], since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

[tex]v=\omega r[/tex]

where

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

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

Answer:

The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex]  times.

Explanation:

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

[tex]\log K=\log A-\frac{Ea}{2.303\times RT}[/tex]

The expression used with catalyst and without catalyst is,

[tex]\log K_1=\log A-\frac{Ea_1}{2.303\times RT}[/tex]...(1)

[tex]\log K_2=\log A-\frac{Ea_2}{2.303\times RT}[/tex]...(2)

On subtracting (2) from (1)

[tex]\log \frac{K_2}{K_1}=\frac{Ea_1-Ea_2}{2.303RT}[/tex]

where,

[tex]K_2[/tex] = rate of reaction with catalyst

[tex]K_1[/tex] = rate of reaction without catalyst  

[tex]Ea_2[/tex] = activation energy with catalyst  = 29.3 kJ/mol = 29300 J/mol

[tex]Ea_1[/tex] = activation energy without catalyst  = 75.3 kJ/mol=75300 J/mol

R = gas constant =8.314 J /mol K

T = temperature = [tex]20^oC=273+20=293K[/tex]

Now on substituting all the values in the above formula, we get

[tex]\log \frac{K_2}{K_1}=\frac{75300 kJ/mol-29300 kJ/mol}{2.303\times 8.314 J/mol K\times 293}=1.58\times 10^{8}[/tex]

The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex]  times.

Explanation:

It is given that,

Mass of the bullet, m₁ = 0.093 kg

Initial speed of bullet, u₁ = 100 m/s

Mass of block, m₂ = 2.5 kg

Initial speed of block, u₂ = 0

We need to find the speed of the block after the bullet embeds itself in the block. Let it is given by V. On applying the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]

[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

[tex]V=\dfrac{0.093\ kg\times 100\ m/s+0}{(0.093\ kg+2.5\ kg)}[/tex]

V = 3.58 m/s

So, the speed of the bullet is 3.58 m/s. Hence, this is the required solution.

Answer:

0.14 J

Explanation:

Use the force to calculate the spring constant.

F = k Δx

27 N = k (0.32 m − 0.28 m)

k = 675 N/m

Work is the change in energy:

W = PE

W = ½ k (Δx)²

W = ½ (675 N/m) (0.30 m − 0.28 m)²

W = 0.135 Nm

W = 0.135 J

Rounding to two decimal places, W = 0.14 J.

Your answer was correct, but it was in units of Ncm, and you needed to answer in units of J.

The work done to stretch the spring from 28 cm to 30 cm is 39.00 Joules, computed using the principles of Hooke's Law and the concept of work done.

In this problem, we are dealing with the concept of work done on a spring, which falls under Physics principles. Hooke's Law states that the force to compress or extend a spring by a distance x from its natural length is proportional to x. It can be written as F = kx, where F is the force, k is the spring constant, and x is the distance.

In this case, the force (F) is 27 N, and the length of stretch (x) is 32 cm - 28 cm = 4 cm. We can find the spring constant (k) using the formula k = F / x = 27 N / 4 cm = 6.75 N/cm.

The work done (W) to stretch the spring from 28 cm to 30 cm is the area under the force/displacement graph from '28 cm' to '30 cm'. Since the force is linear with displacement for a spring, this area can be found using the formula for the area of a trapezoid: W = ½ (F1 + F2) x d. F1 is the initial force (k*28 cm), F2 is the final force (k*30 cm), and d is the displacement (30 cm - 28 cm). Substituting the values, W = ½ [(6.75 N/cm*28 cm)+(6.75 N/cm*30 cm)]*(2 cm) = 39.00 J.

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Answer:

a)

1.73 m/s

b)

6.43 x 10⁻⁴ m

Explanation:

m = mass of the flea = 1.8 x 10⁻⁴ kg

v₀ = initial speed of the flea = 0 m/s

v = final speed of the flea

W = work done by the force on the flea = 2.7 x 10⁻⁴ J

Using work-change in kinetic energy, Work done is given as

W = (0.5) m (v² - v₀²)

Inserting the values

2.7 x 10⁻⁴ = (0.5) (1.8 x 10⁻⁴) (v² - 0²)

v = 1.73 m/s

b)

d = distance moved by the flea while pushing off

F = Upward force applied on the flea by ground = 0.42 N

Work done is also given as

W = F d

2.7 x 10⁻⁴ = (0.42) d

d = 6.43 x 10⁻⁴ m

Answer:

Magnetic field = 1.41 T

Explanation:

Magnetic field of solenoid, B = μnI

μ = 4π x 10⁻⁷N/A²

Number of turns per meter, [tex]n=\frac{N}{L}=\frac{5601}{9.1\times 10^{-2}}=6.15\times 10^4turns/m[/tex]

Current, i = 18.2 A

B = μnI = 4π x 10⁻⁷ x 6.15 x 10⁴ x 18.2 = 1.41 T

Magnetic field = 1.41 T

Answer:

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Answer:

The average normal stress is 5 ksi.

Explanation:

Given that,

Diameter = 2 inch

Load = 15.71 kip

We need to calculate the average normal stress

Using formula of stress

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

Where, F = load

A = area

Put the value into the formula

[tex]\sigma=\dfrac{15.71}{\pi\times(\dfrac{2}{2})^2}[/tex]

[tex]\sigma = 5\ kip/inc^2[/tex]

[tex]\sigma=5\ ksi[/tex]

Hence, The average normal stress is 5 ksi.

Answer:

The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

Explanation:

Given that,

Velocity of car relative to ground = 15.3 m/s

Velocity of truck relative to car = 22.5 m/s

We need to calculate the magnitude of the truck's velocity relative to the ground

We need to calculate the x component of the velocity

[tex]v_{x}=22.5\cos\theta[/tex]

[tex]v_{x}=22.5\cos52^{\circ}[/tex]

[tex]v_{x}=13.852\ m/s[/tex]

We need to calculate the y component of the velocity

[tex]v_{y}=15.3+22.5\sin\theta[/tex]

[tex]v_{y}=15.3+22.5\sin52^{\circ}[/tex]

[tex]v_{y}=33.030\ m/s[/tex]

Using Pythagorean theorem

[tex]|v|=\sqrt{v_{x}^2+v_{y}^2}[/tex]

[tex]|v|=\sqrt{(13.852)^2+(33.030)^2}[/tex]

[tex]|v|=35.82\ m/s[/tex]

Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

Answer:

B) 1218

Explanation:

N = Total number of turns in the solenoid

L = length of the solenoid = 34.00 cm = 0.34 m

B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T

i = current carried by the solenoid = 2.000 A

Magnetic field at the center of the solenoid is given as

[tex]B = \frac{\mu _{o}N i}{L}[/tex]

[tex]9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}[/tex]

N = 1218

The value of N is about B) 1218

[tex]\texttt{ }[/tex]

Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:

[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]

B = magnetic field strength from current carrying wire (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

d = distance (m)

[tex]\texttt{ }[/tex]

[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]

B = magnetic field strength at the center of the solenoid (T)

μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)

I = current (A)

N = number of turns

L = length of solenoid (m)

Let's tackle the problem now !

[tex]\texttt{ }[/tex]

Given:

Current = I = 2000 A

Length = L = 34.00 cm = 0.34 m

Magnetic field strength = B = 9000 mT = 9 T

Permeability of free space = μo = 4π × 10⁻⁷ T.m/A

Asked:

Number of turns = N = ?

Solution:

[tex]B = \mu_o \frac{I N}{L}}[/tex]

[tex]\frac{I N}{L} = B \div \mu_o[/tex]

[tex]IN = BL \div \mu_o[/tex]

[tex]N = BL \div (\mu_o I)[/tex]

[tex]N = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )[/tex]

[tex]\boxed {N \approx 1218}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Magnetic Field