Two skiers travel two different frictionless paths. The two paths start at the same place and finish at the same place. Skier A is heavier than Skier B. With no more information, which of the skiers has the greater speed at the finish? Ignore air resistance.

A. Skier A has the same speed as Skier B at the finish.

B. Skier A has the greater speed at the finish.

C. Skier B has the greater speed at the finish.

D. There is no way to know which skier has the greater speed at the finish.

To solve this problem it is necessary to apply the concepts related to the concepts to continuity.

From the continuity equation we know that

[tex]A_1V_1 = A_2 V_2[/tex]

Where,

A = Cross-sectional Area

V = Velocity

In a circle the area is given by

[tex]A_1 = \pi r_1^2\\A_2 = \pi r_2^2[/tex]

Therefore replacing with our values we have that

[tex]A_1 = \pi r_1^2\\A_1 = \pi 1.8^2\\A_1 = 10.278m^2[/tex]

While the area two is defined as

[tex]A_2 = \pi r_2^2\\A_2 = \pi 0.6^2\\A_2 = 1.13m^2[/tex]

Applying the continuity equation we have to

[tex]A_1V_1 = A_2 V_2[/tex]

[tex](10.278)(3)= (1.13) V_2[/tex]

[tex]V_2 = 27.28m/s[/tex]

Therefore the speed of water flow in the test section is 27.28m/s

Explanation:

Given that,

Distance between two long wires, d = 16 cm = 0.16 m

Current in one wire, [tex]I_1=2.9\ A[/tex]

Current in wire 2, [tex]I_2=5.7\ A[/tex]    

The magnetic force per unit length of one wire on the other is given by the following expression as :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}[/tex]

[tex]\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.9\times 5.7}{2\pi \times 0.16}[/tex]

[tex]\dfrac{F}{l}=2.06\times 10^{-5}\ N/m[/tex]

The current is flowing in opposite direction, the magnetic force acting on it is repulsive. Hence, this is the required solution.

Answer:

5 m/s

Explanation:

mass of bullet, m = 0.5 kg

initial velocity of bullet is u.

mass of pendulum, M = 9.50 kg

height raised, h = 1.28 m

The kinetic energy of the bullet and pendulum system is equal to the potential energy of the system

Let vf be the final velocity after the impact

KE = PE

1/2 (M + m) Vf^2 = (M + m) x g x h

vf^2 = 2 x 9.8 x 1.28 = 25.088

vf = 5 m/s

Thus, the final speed of the combined mass system after the impact s 5 m/s.  

Final answer:

The final speed of the combined masses after the collision can be found using the conservation of momentum and mechanical energy principles, resulting in a final speed of approximately 1.20 m/s.

Explanation:

The final speed Vf of the combined masses after the collision can be calculated using the principle of conservation of momentum and conservation of mechanical energy.

By equating the initial momentum of the system before the collision to the final momentum of the system after the collision, you can determine the final speed Vf.

In this case, the final speed Vf is approximately 1.20 m/s.

To create a gauge pressure of 62.5 cm H₂O on an area of 51 cm², you must exert a force of approximately 31.27 newtons on the balloon. This is obtained by converting the pressure to pascals, the area to square meters and using the formula Pressure = Force/Area.

To find the force you must exert on the balloon to create a gauge pressure of 62.5 cm H₂O when squeezing an area of 51 cm², you can use the relationship between pressure, force, and area given by the formula Pressure (P) = Force (F) / Area (A). First, convert the gauge pressure from cm H₂O to pascals (Pa) since 1 cm H₂O is approximately 98.0665 Pa. Then multiply the converted pressure by the area in m² to find the force in newtons (N).

Step 1: Convert cm H₂O to pascals (Pa):
62.5 cm H₂O * 98.0665 Pa/cm H₂O = 6131.65 Pa

Step 2: Convert the area from cm²to m² (since there are 10,000 cm2 in 1 m²):
51 cm² / 10,000 = 0.0051 m²

Step 3: Calculate the force:
Force (F) = Pressure (P) * Area (A)
F = 6131.65 Pa * 0.0051 m² = 31.27 N

Therefore, you must exert a force of approximately 31.27 newtons on the balloon.

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

[tex]F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N[/tex]

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

[tex]W = K_e[/tex]

[tex]dF_f = 0.5Mv_0^2 [/tex]

[tex]d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m[/tex]

j.Kinetic energy of the bullet before the impact:

[tex]K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J[/tex]

Kinetic energy of the block-bullet system after the impact:

[tex]K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J[/tex]

So 14520 - 33 = 14487 J was lost during the lodging process.

Answer:

therefore, the probability that an individual will have a cholesterol level greater than 60 ​mg/dL.= 0.27

Explanation:

given data

Normal distribution

mean cholesterol level μ= 51.6 ​mg/dL

Standard deviation σ=  14.3 ​mg/dL

x= 60 mg/dL

We have to find out P(x>60)

We Know that [tex]P(x>a) =P(Z>(a-\mu)/\sigma)[/tex]

therefore, [tex]P(x>60) =P(Z>(60-51.3)/14.3)[/tex]

= P(Z>0.61)

= 1 - P(Z<0.61)

= 1 - 0.7291

= 0.27

therefore, the probability that an individual will have a cholesterol level greater than 60 ​mg/dL.= 0.27

To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we calculate the z-score and find the corresponding probability using a Z-table or calculator.

To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we need to use the z-score formula. The z-score is calculated by subtracting the mean from the value of interest and dividing it by the standard deviation. In this case, the z-score is (60 - 51.6) / 14.3 = 0.589. Using a Z-table or a calculator, we can find that the probability of a z-score greater than 0.589 is approximately 0.279.

https://brainly.com/question/34741155

#SPJ12

Answer:

DC: 60V

AC: 15Vrms (21.2V amplitude)

Explanation:

Since we know the current values for each case, we can calculate voltage in both scenarios.

For DC voltage:

Vdc = Idc * R = 60mA * 1000 Ohms = 60Vdc

For AC voltage:

Vac = Iac * R = 15mA * 1000 Ohms = 15 Vac (rms) the amplitude of this signal would be Vmax = 15Vac * 1.414 = 21.2V

Answer:

2.27m/s

Explanation:

The coefficient of sliding friction = 0.38, the incline angle = 60o

Coefficient of sliding friction = frictional force/ force of normal (fn)

The normal force = mgcos60 where m is the mass of the body and g is acceleration due to gravity

Frictional force = coefficient of friction*fn = 0.38*9.81*m*cos 60 = 1.8639m

Force tending to push the body down the incline plane = mgsin60 = 8.496m

Net force on the body = force pushing downward - Frictional force

Net force = 8.496m - 1.8639m = 6.632m

Kinetic energy of the body = 1/2 mv^2 = work done by the net force pushing the body down = net force * distance travelled

1/2mv^2 = 6.632m * 0.387in meters

Cancel mass on both side leaves 1/2v^2 = 6.632*0.387

V^2 = 2.566584*2 = 5.133

V = √5.133 = 2.27m/s

The speed of the block after it has traveled a distance of 38.7 cm is 1.2 m/s.

The speed of the block is calculated by applying the following formula.

Using work energy theorem;

The work done by the force of friction is equal to the change in kinetic energy of the box.

W = K.E

μmgd cosθ  = ¹/₂mv²

μgdcosθ = ¹/₂v²

v² = (2gd cosθ)μ

v = √ (2gd cosθμ)

where;

v = √ (2gd cosθμ)

v = √ (2 x 9.8 x 0.387  x cos(60) x 0.38)

v = 1.2 m/s

Learn more about speed of sliding blocks here: https://brainly.com/question/30440096

#SPJ3

Answer:

Induced EMF,[tex]\epsilon=0.0143\ volts[/tex]

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, [tex]B_i=200\ mT=0.2\ T[/tex]

Final magnetic field, [tex]B_f=300\ mT=0.3\ T[/tex]

The expression for the induced emf is given by :

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

[tex]\phi[/tex] = magnetic flux

[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]

[tex]\epsilon=A\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=A\dfrac{B_f-B_i}{t}[/tex]

[tex]\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}[/tex]

[tex]\epsilon=0.0143\ volts[/tex]

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

Final answer:

The magnitude of the induced emf in the loop is -0.0184 V.

Explanation:

The magnitude of the emf induced in a single-turn circular loop of wire can be calculated using Faraday's law of electromagnetic induction. According to the law, the magnitude of the induced emf is equal to the rate of change of magnetic flux through the loop. The formula for calculating the induced emf is:

emf = -N * ΔB/Δt

Where emf is the induced electromotive force, N is the number of turns in the loop, ΔB is the change in magnetic field strength, and Δt is the change in time. Plugging in the given values, we have:

emf = -1 * (300 - 200) * 10^-3 / 0.0548

emf = -0.0184 V

Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m =  mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

[tex]|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s[/tex]

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3  (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Answer:m=86.36 kg

Explanation:

Given

mass of tackler [tex]m_T=100 kg[/tex]

initial velocity of Tackler [tex]u_t=4.1 m/s[/tex]

Final velocity of combined system [tex]v=2.2 m/s[/tex]

Let m be the mass of receiver

conserving momentum

[tex]m\times 0+m_t\times u_t=(m+m_t)v[/tex]

[tex]0+100\times 4.1=(100+m)\cdot 2.2[/tex]

[tex]410=(100+m)\cdot 2.2[/tex]

[tex]100+m=186.36[/tex]

[tex]m=86.36 kg[/tex]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

Answer:

The speed of the car was 28 m/s.

Explanation:

Hi there!

The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:

W = ΔKE = final kinetic energy - initial kinetic energy

In this case, the final kinetic energy is zero, then:

W = - initial kinetic energy

Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:

Wfr = initial kinetic energy

The work done by friction is calculated as follows:

Wfr = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction

Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).

Then the friction force can be written as follows:

Fr = m · g · μ

Where:

m = mass of the car.

g =  acceleration due to gravity (9.8 m/s²)

The work done by friction will be:

W = m · g · μ · d

The equation of kinetic energy is the following:

KE =  1/2 · m · v²

Where

m = mass of the car.

v = speed.

Then:

W = KE

m · g · μ · d = 1/2 · m · v²

2 · g · μ · d = v²

2 · 9.8 m/s² · 0.40 · 98 m = v²

v = 28 m/s

The speed of the car was 28 m/s.

q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.

To calculate the potential energy of the system of three charges, we use the formula:

PE = k * q1 * q2 / r12 + k * q2 * q3 / r23 + k * q3 * q1 / r31

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q1, q2, and q3 are the charges of the three point charges (4.10 nC, -2.90 nC, and 2.00 nC, respectively)

r12, r23, and r31 are the distances between the three point charges (20.0 cm, 9.0 cm, and 11.0 cm, respectively)

Plugging in the values, we get:

PE = (8.98755 x 10^9 N·m^2/C^2) * (4.10 x 10^-9 C) * (-2.90 x 10^-9 C) / 0.20 m + (8.98755 x 10^9 N·m^2/C^2) * (-2.90 x 10^-9 C) * (2.00 x 10^-9 C) / 0.09 m + (8.98755 x 10^9 N·m^2/C^2) * (2.00 x 10^-9 C) * (4.10 x 10^-9 C) / 0.11 m

PE = 1.19 x 10^-6 J

Therefore, the potential energy of the system of three charges if q3 is placed at x = + 11.0 cm is 1.19 x 10^-6 J.

To make the potential energy of the system equal to zero, we need to place q3 at a point where the forces from q1 and q2 cancel each other out. This means that the vectors representing the forces must be equal and in opposite directions.

The force from q1 on q3 is given by:

F13 = k * q1 * q3 / r31^2

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q1 is the charge of q1 (4.10 nC)

q3 is the charge of q3 (2.00 nC)

r31 is the distance between q1 and q3

The force from q2 on q3 is given by:

F23 = k * q2 * q3 / r23^2

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q2 is the charge of q2 (-2.90 nC)

q3 is the charge of q3 (2.00 nC)

r23 is the distance between q2 and q3

For the forces to cancel each other out, we need:

F13 = -F23

Substituting in the equations for the forces, we get:

k * q1 * q3 / r31^2 = -k * q2 * q3 / r23^2

Solving for r31, we get:

r31 = r23 * sqrt(q2/q1)

Substituting in the values, we get:

r31 = 0.09 m * sqrt(-2.90 nC / 4.10 nC)

r31 = 0.069 m

Therefore, q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.

for such more question on potential energy

https://brainly.com/question/1455245

#SPJ3

The 93 kg man will be moving at a speed of 0.161 m/s after catching the ball that he threw at a wall and rebounded back to him, based on the principle of conservation of momentum.

This question involves the principle of conservation of momentum. According to the law of conservation of momentum, the total momentum of the system (man and the ball) is conserved before and after the collision with the wall. The total initial momentum, before the ball is thrown, is zero as everything is at rest.

When the man throws the ball, the ball gains momentum in one direction, and to keep the total momentum zero, the man must gain an equal amount of momentum in the opposite direction. On returning, the ball has changed its direction, so it has lost the initial momentum and gained the same amount in the opposite direction.

The man catches the ball, meaning he gains the momentum that the ball brought back, causing him to move. To determine how fast he is moving, we set the final momentum of the system (which should still be zero) equal to the initial momentum.

To calculate the speed (v) of the man we'll use the equation: v = 2 × ball_mass × ball_speed / man_mass = 2 × 0.653 kg × 11.3 m/s / 93 kg = 0.161 m/s

https://brainly.com/question/33316833

#SPJ12

The man will be moving at approximately 0.078 m/s after he catches the ball. This is found using the conservation of momentum, considering the ball's momentum before and after it collides with the wall and bounces back to the man.

The student's question involves the conservation of momentum, a fundamental concept in physics. When the man throws the ball and catches it after it bounces off the wall, both the ball and the man are involved in an isolated system where external forces are negligible. As momentum is conserved, the velocity of the man after catching the ball can be calculated using the conservation of momentum principle.

To solve this problem, we first consider the momentum before and after the collision with the wall. The man throws the ball at 11.3 m/s. The direction of the ball's momentum changes after it bounces back, but the speed remains the same due to the 'elastic' nature of the collision as mentioned in the problem statement (no energy loss).

The initial momentum of the system (man and ball) is zero because they're both at rest initially. When the man throws the ball, the ball gains momentum in the forward direction, and the man gains momentum in the opposite direction. This momentum for the man (in the opposite direction) is what will be calculated.

The initial momentum of the ball and man is:

The final momentum after throwing the ball is:

By conservation of momentum:

Initial momentum = Final momentum

0 = -0.653 kg * 11.3 m/s + 93 kg * v

Solving for v gives:

v = (0.653 kg * 11.3 m/s) / 93 kg

v ≈ 0.078 m/s

This is the speed of the man after catching the ball, moving opposite to the direction in which he threw the ball.

https://brainly.com/question/33316833

#SPJ3

Answer:

 [tex]c_{e1}[/tex] = 128.3 J / kg ° C

Explanation:

In this exercise we will use that the expression for heat is

    Q = m [tex]c_{e}[/tex] ΔT

As they indicate that there are no losses with the medium, the heat transferred by the tungsten is equal to the heat absorbed by the water plus the calorimeter

    Q assigned = QAbsorbed

    Q hot = Q cold + Q calorimeter

The mass of tungsten (m₁ = 69.12 10⁻³ kg) with an initial temperature (T₁ = 98.93°C),

The mass of water (m₂ = 85.45 10⁻³ kg) at a temperature (T₂ = 23.82°C),

a calorimeter constant (C = 1.56 J/ °C)

m₁ [tex]c_{e1}[/tex] (T₁ - [tex]T_{f}[/tex]) = (m₂ [tex]c_{e2}[/tex] + C) ([tex]T_{f}[/tex] - T₂)  

[tex]c_{e1}[/tex]= (m₂ ce2 + C) ([tex]T_{f}[/tex]-T₀) / (m₁ (T₁-[tex]T_{f}[/tex])  

[tex]c_{e1}[/tex] = (85.45 10-3 4186 + 1.56) (25.63 - 23.82) / (69.12 10-3 (98.93 - 25.63))  

[tex]c_{e1}[/tex] = (357.69 + 1.56) 1.81 / (69.12 10-3 73.3)  

[tex]c_{e1}[/tex] = 650.24 / 5.0665  

[tex]c_{e1}[/tex] = 128.3 J / kg ° C

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate [tex]m=100 kg[/tex]

Crate slows down in [tex]s=1.5 m[/tex]

initial speed [tex]u=1.77 m/s[/tex]

inclination [tex]\theta =30^{\circ}[/tex]

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

[tex]W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2[/tex]

[tex]W_{gravity}=mg(0-h)=mgs\sin \theta [/tex]

[tex]W_{gravity}=-mgs\sin \theta[/tex]

[tex]W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N[/tex]

change in kinetic energy[tex]=\frac{1}{2}\times 100\times 1.77^2=156.64 J[/tex]

[tex]W_{friction}=156.64+735=891.645[/tex]

(b)Coefficient of sliding friction

[tex]f_r\cdot s=W_{friciton}[/tex]

[tex]891.645=f_r\times 1.5[/tex]

[tex]f_r=594.43 N[/tex]

and [tex]f_r=\mu mg\cos \theta [/tex]

[tex]\mu 100\times 9.8\times \cos 30=594.43[/tex]

[tex]\mu =0.7[/tex]

Final answer:

To answer part (a), calculate the work done by friction using the equation work = force × distance. Calculate the change in kinetic energy of the crate using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2 and subtract it from the work done by friction. To answer part (b), use the equation frictional force = coefficient of friction × normal force and calculate the normal force using the equation weight = mass × acceleration due to gravity.

Explanation:

To answer part (a), we need to calculate the work done by friction. The work done by friction can be calculated using the equation work = force × distance. In this case, the force is the frictional force and the distance is the distance traveled by the crate. The work done by friction is equal to the change in kinetic energy of the crate, which can be calculated using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2. Subtracting the change in kinetic energy from the work done by friction will give us the energy dissipated by friction.

To answer part (b), we can use the equation frictional force = coefficient of friction × normal force. The normal force is equal to the weight of the crate, which can be calculated using the equation weight = mass × acceleration due to gravity. By substituting the known values into these equations, we can find the coefficient of sliding friction.

Answer:

The number of available energy is [tex]4.820\times10^{45}[/tex]

Explanation:

Given that,

Energy [tex]E=4.9\times10^{-4}\ J[/tex]

Temperature = 2.7 K

Energy states per unit volume [tex]dE=4\times10^{-5}\ eV[/tex]

We need to calculate the number of available energy

Using formula of energy

[tex]N=g(E)dE[/tex]

[tex]N=\dfrac{8\pi\times E^2 dE}{(hc)^3}[/tex]

Where, h = Planck constant

c = speed of light

E = energy

Put the value into the formula

[tex]N=\dfrac{8\pi\times(4.9\times10^{-4})^2\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]

[tex]N=4.820\times10^{45}[/tex]

Hence, The number of available energy is [tex]4.820\times10^{45}[/tex]

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

[tex]\sum F = ma[/tex]

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

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

To find the required force then,

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

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

[tex]F = \mu mg +ma[/tex]

[tex]F = 0.35*50*0.8+50*1.2[/tex]

[tex]F=(171.5N)+(50Kg)(1.2m/s^2)[/tex]

[tex]F=231.5N[/tex]

[tex]F\approx 230N[/tex]

Therefore the horizontal force applied on the block is B) 230N

To find the horizontal force necessary to accelerate the box at 1.2 m/s^2, we consider the frictional forces. The maximum static friction can be calculated using the coefficient of static friction and the weight of the box. The horizontal force required is equal to the maximum static friction.

To find the horizontal force required to accelerate the 50-kg box at 1.2 m/s^2, we need to consider the frictional forces acting on the box. First, we calculate the maximum static friction using the formula:

Maximum static friction = coefficient of static friction × normal force

Normal force = mass × gravity

Maximum static friction = 0.65 × (50 kg × 9.8 m/s^2)

Horizontal force = maximum static friction = 0.65 × (50 kg × 9.8 m/s^2)

So, the horizontal force needed is approximately 318.5 N.

https://brainly.com/question/33794648

#SPJ11

Answer

given,

height of aquarium = 6 m

Depth of fresh water = D = 1.50 m

horizontal length of the aquarium(w) = 8.40 m

total force increased when liquid is filled to depth = 4.30 m

g = 9.81 m/s²

ρ = 998 Kg/m³

force in the aquarium.

dF = PdA

[tex]F = \int PdA[/tex]

[tex]F = \int \rho\ g\ y\ (wdy)[/tex]

[tex]F = \rho\ g\ w \int ydy[/tex]

[tex]F = \rho\ g\ w\dfrac{y^2}{2}[/tex]

[tex]F = \dfrac{\rho\ g\ w\ y^2}{2}[/tex]

At D = 1.5 m

[tex]F = \dfrac{980\times 9.8\times 8.4\times 1.5^2}{2}[/tex]

F = 9.08 x 10⁴ N

At D = 4.30 m

[tex]F = \dfrac{980\times 9.8\times 8.4\times 4.3^2}{2}[/tex]

F = 7.46 x 10⁵ N

Total force on the wall increased by

ΔF  = 74.6 x 10⁴ - 9.08 x 10⁴

ΔF  = 65.52 x 10⁴ N

To find the heat required to melt the ice on the windshield, calculate the mass of the ice using its thickness, area, and density, then use the formula for heat transfer to find the heat required using the latent heat of fusion for ice.

In this Physics problem, we're trying to calculate the heat needed to melt the -4.0 °C ice from a windshield. The heat required to melt ice is given by the formula Q = m * Lf, where Q is the heat, m is the mass of the substance (ice in this case), and Lf is the latent heat of fusion, which is the amount of heat necessary to change the state of the substance from solid to liquid at its melting point.

To calculate the mass of the ice, we use the density (ρ) given and the volume of the ice, which can be found by multiplying the thickness of the ice by the area of the windshield and converting cm to m. Volume of the ice would be Area * thickness = 1.0 m^2 * 0.007 m (from 0.7 cm) = 0.007 m^3. Plug this into ρ = m/V to get the mass m = ρ * V = 917 kg/m^3 * 0.007 m^3 = 6.42 kg.

Finally, we use Q = m * Lf which is Q = 6.42 kg * 334 kJ/kg (given that the heat of fusion for water is 334 kJ/kg), we get Q = 2143 kJ required to melt all the ice.

https://brainly.com/question/13433948

#SPJ12

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

[tex]n_{r}=1.572[/tex]

[tex]n_{b}=1.587[/tex]

We need to calculate the angle for red wavelength

Using Snell's law,

[tex]n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}[/tex]

Put the value into the formula

[tex]1.572\sin37=1\times\sin\theta_{r}[/tex]

[tex]\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})[/tex]

[tex]\theta_{r}=71.0^{\circ}[/tex]

We need to calculate the angle for blue wavelength

Using Snell's law,

[tex]n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}[/tex]

Put the value into the formula

[tex]1.587\sin37=1\times\sin\theta_{b}[/tex]

[tex]\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})[/tex]

[tex]\theta_{b}=72.7^{\circ}[/tex]

We need to calculate the angle between the red and blue light

Using formula of angle

[tex]\Delta \theta=\theta_{b}-\theta_{r}[/tex]

Put the value into the formula

[tex]\Delta \theta=72.7-71.0[/tex]

[tex]\Delta \theta=1.7^{\circ}[/tex]

Hence, The angle between the red and blue light is 1.7°.

The Hydrogen discharge lamp emits light of specific wavelengths which pass through a flint-glass prism, refract (or bend), and display a color spectrum due to dispersion. The red and blue lights have different refraction angles because of their different wavelengths. The distinct color bands indicate discrete energy transitions in the hydrogen gas.

Light emission and its properties are central to this question. The light emitted by a hydrogen discharge lamp, which contains hydrogen gas at low pressure, indicates that there are discrete energies emitted, thus producing light of specific wavelengths. These distinct wavelengths are a result of H2 molecules being broken apart into separate H atoms when an electric discharge passes through the tube.

When this light enters a prism, it refracts or bends, leading to a phenomenon called dispersion. This happens because not all colors or wavelengths of light are bent the same amount: the degree of bending depends on the wavelength of the light, as well as the properties of the material (in this case, flint-glass prism). For instance, red and blue light have different refraction angles due to their different wavelengths of 656 nm and 486 nm respectively.

So, in the context of a hydrogen discharge lamp, the two prominent wavelengths, 656 nm (red) and 486 nm (blue), will refract differently when passing through the flint-glass prism. Further, the separation (or dispersion) of light into these distinct colored bands as it exits the prism gives us a line spectrum - visual evidence of the discrete energy transitions happening at the atomic level in the hydrogen gas of the lamp.

https://brainly.com/question/20259452

#SPJ3

Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, [tex]V_{p} = 120\ V[/tex]          (rms voltage)

Voltage at secondary, [tex]V_{s} = 13000\ V[/tex]  (rms voltage)

Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]  

Now,

(a) The ratio of secondary to primary turns is given by the relation:

[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]

where

[tex]N_{p}[/tex] = No. of turns in primary

[tex]N_{s}[/tex] = No. of turns in secondary

[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108

(b) The power supplied to the line is given by:

Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]

(c) The current rating that the fuse should have is given by:

[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]

[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]

[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]

Answer:

The cheerleader does a total work of 8,532 J (8,541 J without any intermediate rounding)

Explanation:

Hi there!

The equation of work is the following:

W = F · d

Where:

W = work.

F = applied force.

d = distance.

The cheerleader´s partner is on the ground due to the force of gravity acting on her, also known as weight. To lift the partner at a constant velocity, the cheerleader has to suppress this force applying on the partner a force that is equal to the partner´s weight but in opposite direction.

The weight of the partner is calculated as follows:

Weight = m · g

Where m is the mass of the partner and g is the acceleration due to gravity.

Let´s calculate the weight of the partner:

Weight = 43.8 kg · 9.8 m/s²

Weight = 429 N

]Then, the work done in one lift by the cheerleader is the following:

W = F · d

W = 429 N · 0.737 = 316 J

If the cheerleader does 27 lifts, the work done will be (316 J · 27) 8,532 J

The work done by the cheerleader to lift his partner 27 times is calculated as the product of force (which itself is the product of the partner's mass and gravity) and distance, hence resulting in a total work done of 8570.61 Joules.

In Physics, energy exerted to move an object is often calculated as work done. Work is defined as the product of the force applied to an object and the distance over which it is applied, and its unit of measurement is Joules (J). Regarding the given problem, the work done by the cheerleader to lift his partner is calculated using the formula: Work = force x distance, where force is derived from the product of the partner's mass and gravity (Force = mass x gravity). So, the work done for one lift would be calculated as follows:

Work(std.) = Force * distance = mass * gravity * distance = 43.8 kg * 9.8 m/s ² * 0.737 m = 317.43 J.

To find out the total work done after 27 lifts, simply multiply this result by 27: Total Work = 27 * Work(std.) = 27 * 317.43 J = 8570.61 J. Thus, the cheerleader does 8570.61 Joules of work by lifting his partner 27 times.

https://brainly.com/question/35917320

#SPJ11

To solve this problem it is necessary to apply the concepts related to energy as a function of voltage, load and force, and the definition of Force given by Newton in his second law.

By definition we know that force is equal to

F= ma

Where,

m = mass (at this case of an electron)

a = Acceleration

But we also know that the Energy of an electric object is given by two similar definitions.

[tex]1) E= \frac{F}{q}[/tex]

Where,

F= Force

q = Charge of proton/electron

[tex]2) E = \frac{V}{d}[/tex]

V = Voltage

d = Distance

Equating and rearrange for F,

[tex]\frac{F}{q} = \frac{V}{d}[/tex]

[tex]F = \frac{Vq}{d}[/tex]

The two concepts of force can be related to each other, then

[tex]ma = \frac{Vq}{d}[/tex]

Acceleration would be,

[tex]a = \frac{Vd}{dm}[/tex]

Replacing with our values we have that the acceleration is

[tex]a = \frac{Vq}{dm}[/tex]

[tex]a = \frac{(170)(1.6*10^{-19})}{(2*10^{-2})(9.1*10^{-31})}[/tex]

[tex]a = 1.49*10^{15}m/s^2[/tex]

Now through the cinematic equations of motion we know that,

[tex]V_f^2-V_i^2 = 2ax[/tex]

Where,

[tex]V_f =[/tex] Final velocity

[tex]V_i =[/tex] Initial velocity

a = Acceleration

x = Displacement

Re-arrange to find v_f,

[tex]v_f = \sqrt{v_i^2+2ax}[/tex]

[tex]v_f = \sqrt{2.5*10^5+2*(1.49*10^{15})(0.1*10^{-2})}[/tex]

[tex]v_f = 1.726*10^6 m/s[/tex]

Therefore the electron's speed when it is 0.1 cm from the negative plate is [tex]1.726*10^6 m/s[/tex]

Speed of the electron enters in the gape of two large aluminum plates,  when it is 0.1 cm from the negative plate is,

[tex]v_f=1.726\times10^6\rm m/s[/tex]

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

The electric force due to the electric field can be given as,

[tex]F=\dfrac{Vq}{d}[/tex]

Here, (V) is the potential difference, (q) is the charge on the body and (d) is the distance.

The charge on a electron is [tex]1.6\times10^{-19}[/tex] C. As the distance between the two plates is 2 meters and the value of potential difference is 170 V. Thus, put the values in the above formula as,

[tex]F=\dfrac{170\times(1.6\times10^{-19})}{0.02}[/tex]

As the force on a body is the product of mass times acceleration and the mass of the electron is [tex]9.1\times10^{-31}[/tex] kg. Thus, the above equation can be rewrite as,

[tex](9.1\times10^{-31})a=\dfrac{170\times(1.6\times10^{-19})}{0.02}[/tex]

[tex]a=1.49\times10^{15}\rm m/s^2[/tex]

The initial speed of the electron is [tex]2.5\times10^5[/tex] m/s and the acceleration is [tex]1.49\times10^{15}\rm m/s^2[/tex]. Thus from the third equation of motion, the final velocity can be given as,

[tex]v_f=\sqrt{2.5\times10^5+2(1.49\times10^{15})(0.1\times10^{-2})}\\v_f=1.726\times10^6\rm m/s[/tex]

Thus, the speed of the electron,  when it is 0.1 cm from the negative plate is,

[tex]v_f=1.726\times10^6\rm m/s[/tex]

Learn more about electric field here;

https://brainly.com/question/14372859

Answer:

605447.7066 kgm²/s

Explanation:

[tex]m_1[/tex] = Mass of sphere = 10000 kg

[tex]m_2[/tex] = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

[tex]\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s[/tex]

Angular momentum is given by

[tex]L=I\omega[/tex]

Moment of inertia of the satellite is

[tex]I_s=\frac{2}{5}m_1r^2[/tex]

Moment of antenna of the satellite is

[tex]I_a=\frac{1}{3}m_2l^2[/tex]

The angular momentum of the system is

[tex]L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s[/tex]

The angular momentum of the satellite is 605447.7066 kgm²/s

The angular momentum of a rotating satellite can be calculated by considering the moment of inertia of its components, including the main body and antennas. The moment of inertia is calculated based on the mass and distribution of mass in each component. The total angular momentum is the sum of the angular momentum of the main body and the angular momentum of the antennas.

The angular momentum of a rotating object can be calculated by multiplying its moment of inertia with its angular velocity. In this case, the satellite consists of a main body in the shape of a sphere and two antennas projecting out from the center of mass. The moment of inertia can be calculated by considering the mass and distribution of mass in each component of the satellite. The angular momentum is the sum of the angular momentum of the main body and the angular momentum of the antennas.

For the main body of the satellite, the moment of inertia can be calculated as: I = (2/5)×m × r², where m is the mass and r is the radius of the sphere. For the antennas, the moment of inertia can be calculated as: I = (1/12) × m × L², where m is the mass and L is the length of each antenna.

Finally, the total angular momentum can be calculated by summing the angular momentum of the main body and the angular momentum of the antennas: L_total = L_main body + 2× L_antennas, since there are two antennas.

https://brainly.com/question/37906446

#SPJ3

Answer:

4.5 x 10^6 miles

Calculations can be viewed on the snapshot attached to this reply.

Thanks

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

Let the mass of each ball be m kg

v[tex]_{1}[/tex] be the velocity of ball A along positive x axis

v[tex]_{2}[/tex] be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v[tex]_{1}[/tex]

∴  v[tex]_{1}[/tex] = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v[tex]_{2}[/tex]

2 = u +  v[tex]_{2}[/tex] → equation 1

∴ relative velocity of approach = relative velocity of separation

-2 =  v[tex]_{2}[/tex] - u → equation 2

By adding both equations 1 and 2 we get

v[tex]_{2}[/tex] = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

In order to solve this problem, it is necessary to apply the concepts related to the Pressure according to the Force and the Area as well as to the pressure depending on the density, gravity and height.

In the first instance we know that the pressure can be defined as

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

Where

F= Force

A =  Mass

In the second instance the pressure can also be defined as

[tex]P = \rho gh[/tex]

Where,

[tex]\rho=[/tex]Density of Fluid at this case Water

g = Gravitational Acceleration

h = Height

If we develop the problem to find the pressure then,

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

[tex]P = \frac{F}{\pi r^2}[/tex]

[tex]P = \frac{2*10^6}{\pi (0.1)^2}[/tex]

[tex]P = 6.37*10^{7} Pa[/tex]

Through the second equation we can find the depth to which it can be submerged,

[tex]P = \rho gh[/tex]

Re-arrange to find h

[tex]h = \frac{P}{\rho g}[/tex]

[tex]h = \frac{6.37*10^{7}}{1000*9.8}[/tex]

[tex]h = 6499.4m[/tex]

The maximum mass of phosphoric acid that can be formed is 34.2 g.

The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex]

The excess reagent that remains after the reaction is complete is 3.76 g.

Phosphoric acid (H3PO4), also known as orthophosphoric acid, is the most important oxygen acid of phosphorus.

It is a crystalline solid and is used to generate fertilizer phosphate salts.

Given,

The balanced chemical equation is

[tex]P_4O_1_0 + 6 H_2O =4 H_3PO_4[/tex]

The molar mass of the following

[tex]\bold{P_4O_1_0}[/tex] = 284 g

[tex]\bold{H_2O}[/tex] = 16 g

[tex]\bold{H_3PO_4}[/tex] = 98 g

Calculating the number of moles

[tex]\dfrac{\bold{P_4O_1_0}}{284 g} = 0.0873\; mol/P_4O_1_0[/tex]

[tex]\dfrac{\bold{13.2\;g\;H_2O}}{18 g} = 0.733\; mol/P_4O_1_0[/tex]

From equation, 1 mol of [tex]\bold{P_4O_1_0}[/tex] reacts with 6 mol of water, then for 0.0873 mol of [tex]\bold{P_4O_1_0}[/tex] will react

[tex]0.0873\; mol\; P_4O_1_0 \times 6 mol H_2O/ 1 molP_4O_1_0 = 0.524\; mol\; H_2O[/tex]

The number of moles of water remaining are:

0.733 mol - 0.524 mol = 0.209 mol

Convert the amount of water into mass unit

0.209 mol · 18 g / mol = 3.76 g

The number of moles of water remaining is 3.76 g

Assuming that the reaction has a yield of 100%

Since 1 mol of [tex]\bold{P_4O_1_0}[/tex]  produces 4 mol of phosphoric acid, 0.0873 mol [tex]\bold{P_4O_1_0}[/tex] will produce:

[tex]0.0873\; mol\; P_4O_1_0 \times 4 mol H_3PO_4/ molP_4O_1_0 = 0.349\; mol\; of\;phosphoric\;acid[/tex]

Then, the maximum mass of phosphoric acid that can be formed is

[tex]0.349\; mol\; H_3PO_4 \times 98\;g/ mol = 34.2\; g[/tex]

Thus, The maximum mass of phosphoric acid is 34.2 g, The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex], and the excess reagent that remains after the reaction is complete is 3.76 g.

Learn more about phosphoric acid, here

https://brainly.com/question/8156365

Final answer:

The maximum mass of phosphoric acid that can be formed is 35.6 grams, with diphosphorus pentoxide being the limiting reagent and 8.77 grams of water remaining as excess.

Explanation:

The maximum mass of phosphoric acid that can be formed is 35.6 grams. The formula for the limiting reagent is diphosphorus pentoxide (P4O6). After the reaction is complete, 8.77 grams of water will remain as the excess reagent.

Usually, to find the maximum mass of the product formed (phosphoric acid), and the mass of the excess reagent that remains after the reaction, it involves determining the limiting reagent by calculating the moles of reactants and using stoichiometry. The limiting reagent is the reactant that will be completely used up first during the chemical reaction, limiting the amount of product formed.