0.140-kg baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove

Answer:

The answer to the question is;

When she is 8 m from the building fast the length of her shadow on the building is decreasing at [tex]\frac{2}{9} m/s[/tex] or 0.22 m/s.

Explanation:

We have

Distance of the spotlight from the building = 20 m

Distance  of woman from the building when her speed is measured = 8 m

Height of the woman = 2 m

Actual speed of the woman = 0.8 m/s

Comparing the distance of the woman from the spotlight and the wall from the spotlight, we have when the woman is 8 m from the building she is 12 m from the spotlight    

Therefore we have

[tex]\frac{12}{20} = \frac{2}{y}[/tex] where y is the shadow cast by the woman on the building = 10/3

When the woman is x distance from the building, she is 20 - x meters from the spotlight

Therefore the above equation can be written  as

[tex]\frac{20-x}{20} = \frac{2}{y}[/tex]  which gives [tex]1 - \frac{1}{20}*x = 2* \frac{1}{y}[/tex] finding the derivative of both sides gives

[tex]-\frac{1}{20}dx =-2*\frac{1}{y^2}dy[/tex] hence we have by dividing by dt gives [tex]-\frac{1}{20}\frac{dx}{dt} =-2*\frac{1}{y^2}\frac{dy}{dt}[/tex]

However we know that [tex]\frac{dx}{dt} = 0.8 m/s[/tex]

Therefore [tex]-\frac{0.8}{20} = -0.18\frac{dy}{dt}[/tex]

The rate of decrease of her shadow [tex]\frac{dy}{dt}[/tex] is given by

[tex]\frac{dy}{dt} = \frac{0.8}{3.6} =\frac{2}{9} m/s[/tex] or 0.222 m/s.

Explanation:

The best methods for the transfer of heat are conduction convection, radiation and sometimes evaporation also.

Answer:Condution radiation convection

Explanation:

Answer:7.67

Explanation:

Given

Weight of Cathy [tex]W_1=460\ N[/tex]

Force exerted by rope [tex]F=60\ N[/tex]

Mechanical advantage is the ratio of load by Pulling effort      

[tex]M.A.=\frac{Load}{Pulling\ effort}[/tex]

[tex]M.A.=\frac{460}{60}[/tex]

[tex]M.A.=7.67\ N[/tex]

Final answer:

The actual mechanical advantage of the pulley system that hoists Cathy is calculated by dividing the load force (460 N) by the effort force (60 N), resulting in an actual mechanical advantage of approximately 7.67.

Explanation:

The actual mechanical advantage (MA) of a pulley system is the ratio of the load force to the effort force. Given that Cathy, a 460-N actress, is raised with an effort of 60 N, we can calculate the actual mechanical advantage by using the formula MA = Load/Effort. By dividing 460 N by 60 N, we get an actual mechanical advantage of approximately 7.67. This implies that the pulley system is using 7.67 times less effort force to lift the load, thus increasing the efficiency of the task.

Pulley systems can provide significant mechanical advantage by allowing a smaller effort force to move a larger load. This effectiveness is heightened if the pulleys are arranged in such a way as to multiply the tension in the ropes as they support the load. With perfectly friction-free pulleys and ropes, the MA can be simply counted as the number of ropes supporting the load, making the force output nearly an integral multiple of the input force.

Answer:

0.00970 s

Explanation:

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

Force due to magnetic field = qvB sin θ

q = charge on the particle = 5.4 μC

v = velocity of the charge

B = magnetic field strength = 2.7 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

where w = angular velocity.

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

mvw = qvB

mw = qB

w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)

w = 3.24 × 10² rad/s

w = 324 rad/s

w = (angular displacement)/time

Time = (angular displacement)/w

Angular displacement = π rads (half of a circle; 2π/2)

Time = (π/324) = 0.00970 s

Hope this Helps!!!

Answer:

2.8 rad/s

Explanation:

In absence of external forces, the total angular momentum of the system must be conserved.

The angular momentum when the arms of the student are extended horizontally is given by:

[tex]L_1 = (I_0 + 2I)\omega_1[/tex]

where:

[tex]I_0=3.8 kg m^2[/tex] is the moment of inertia of the student+stool

[tex]I=mr^2[/tex] is the moment of inertia of each mass, where:

m = 2.6 kg is one mass

r = 0.71 m is the distance of each mass from the rotation axis

[tex]\omega_1=1.8 rad/s[/tex] is the initial angular velocity

So we have

[tex]L_1=(I_0+2mr^2)\omega_1[/tex]

When the student pulls the weights to a distance of r' = 0.23 m, the angular momentum is:

[tex]L_2=(I_0+2I')\omega_2[/tex]

where:

[tex]I'=mr'^2[/tex] is the new moment of inertia of each mass, with

r' = 0.23 m

Since the angular momentum must be constant, we have:

[tex]L_1=L_2\\(I_0+2mr^2)\omega_1 = (I_0+2mr'^2)\omega_2[/tex]

and solving for [tex]\omega_2[/tex], we find the final angular velocity:

[tex]\omega_2 = \frac{I_o+2mr^2}{I_0+2mr'^2}\omega_1=\frac{3.8+2(2.6)(0.71)^2}{3.8+2(2.6)(0.23)^2}(1.8)=2.8 rad/s[/tex]

Answer:

The correct option is

(e)either (c) or (d) could be correct.

Explanation:

The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.

Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since

Work = Force × distance = 0 where distance ≠ 0.

The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.

Therefore it is either horizontally to the left, or horizontally to the right.

Answer:

the ball takes 1.53 s to reach the top of its trajectory.

Explanation:

given information:

the speed, v = - 15 m/s (moving upward)

(a) How much time does it take for the ball to reach the top of its trajectory?

we know that the speed for the vertical motion is

v = v₀ - gt, v₀ = 0

where

v = speed (m/s)

g = gravitational force (9.8 m/s²)

t = time (s)

thus

v = - gt

-15 = - 9.8 t

t = 15/9.8

 = 1.53 s

so, the time that is needed by the ball to reach the top its trajectory is 1.53 s

Answer:

The ball takes 1.53 seconds to reach its top trajectory

Explanation:

The velocity of the ball will keep pushing it upwards until the velocity becomes zero. Therefore, the ball will reach the top of its trajectory when velocity i.e. V=0,

Fundamental principal of velocity is V = Vo + g*t

where, V=0  

Vo = 15 m/s

g = -9.8 m/s^2 (since ball is going upwards against the gravity)

t = ?  

0 = 15 + (-9.8 * t)

-15 = -9.8t

-15 / -9.8 = t

t = 1.53 Seconds

The ball takes 1.53 seconds to reach its top trajectory

Answer:

never, current is not induced

Explanation:

The induced emf in the ring equals the rate of change of magnetic flux in the ring.

E = -dФ/dt = -dAB/dt = -BdA/dt since B the magnetic field is constant.

E = -BdA/dt

Now dA/dt = dA/dy × dy/dt where dA/dy = rate of change of area with vertical distance as the ring is falling. dy/dt = speed of ring.

Since the ring is falling freely before being caught, its speed v is gotten from v = u + at where u = 0 and a = -g

v = 0 - gt = -gt

v = -gt  

So, dA/dt = dA/dy × dy/dt = vdA/dy = -gtdA/dy

So E = -BdA/dt = -B × -gtdA/dy = BgtdA/dy

Since dA/dy = 0 since the area of the ring does not change with vertical distance. So,

E = BgtdA/dy = Bgt × 0 = 0

E = 0

So, emf is never induced because the flux through the ring remains constant

Let us consider the object to be a cell.

Explanation:

As a cell grows bigger, its internal volume enlarges and the cell membrane expands. This results in the increase in volume more rapidly than does the surface area, and so the relative amount of surface area available to pass materials to a unit volume of the cell steadily decreases. This means the surface area to the volume ratio gets smaller as the cell gets larger.

Example of a cube:

Cube size            Surface area                   Volume

2cm                      2 × 2 × 6 = 24 cm²         2 × 2 × 2 = 8 cm³

4 cm                      4 × 4 × 6 = 96 cm²         4 × 4 × 4 = 64 cm³

6 cm                      6 × 6 × 6 = 216 cm²       6 × 6 × 6 = 216 cm³

8 cm                      8 × 8 × 6 = 384 cm²       8 × 8 × 8 = 512 cm³

This shows as the object gets larger, the volume of an object increases faster than the surface area.

Answer:

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.

Answer:

D. Battery

During the oxidation-reduction(redox) reaction, there is always flow of electrons from one point to another. The electrons are then converted to power through the battery which converts chemical energy to electrical energy. If there is zero flow of electrons then there will also be zero power.

Answer:

D. A battery.

Explanation:

A battery cell refers to a single anode and cathode separated by electrolyte used to produce a voltage and current. It is typically an electrolytic cell.

An electrolytic cell is an electrochemical cell that drives a non-spontaneous redox reaction through the application of electrical energy. They are often used to decompose chemical compounds, in a process called electrolysis.

The anode which is positive electrode undergoes oxidation i.e loss of electrons while the cathode(negative) undergoes reduction that is, accept electrons.

Answer:

[tex]h = 2.821\,m[/tex]

Explanation:

The speed of the actor before the collision is found by means of the Principle of Energy Conservation:

[tex](83\,kg)\cdot(9.807\,\frac{m}{s})\cdot (3.90\,m) = \frac{1}{2}\cdot (83\,kg)\cdot v^{2}[/tex]

[tex]v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (3.90\,m)}[/tex]

[tex]v \approx 8.746\,\frac{m}{s}[/tex]

The speed after the inelastic collision is obtained by using the Principle of Momentum Conservation:

[tex](83\,kg)\cdot (8.746\,\frac{m}{s} )+(55\,kg)\cdot (0\,\frac{m}{s} ) = (83\,kg + 55\,kg)\cdot v[/tex]

[tex]v = 5.260\,\frac{m}{s}[/tex]

Lastly, the maximum height is determined by using the Principle of Energy Conservation again:

[tex]\frac{1}{2}\cdot (138\,kg)\cdot (5.260\,\frac{m}{s} )^{2} = (138\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot h[/tex]

[tex]h = \frac{(5.260\,\frac{m}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]h = 2.821\,m[/tex]

Final answer:

Bending the knees when landing from a jump increases the impact time and decreases the force experienced by the body, as per the impulse-momentum theorem. This technique prevents injuries by distributing the force over a longer period, thus reducing the force's intensity on the bones.

Explanation:

When we jump from a height to the ground and bend our knees upon landing, we are effectively reducing the impact force experienced by our body. This is explained by the impulse-momentum theorem, which states that the change in momentum (or impulse) of an object is equal to the force applied multiplied by the time the force is applied. The momentum of a person jumping from a height is quite large due to their mass and the speed at which they are moving toward the ground due to gravity. If we land with straight legs, that momentum is transferred very quickly to our legs and the ground, which results in a very high force over a short time. This can lead to injuries as bones can fracture if the force on them is too large.

By bending our knees, we increase the time over which our momentum is brought to zero, which means the force exerted on our legs and subsequently on the bones is lessened, as force is inversely proportional to the time over which the momentum changes. This principle applies to various real-world situations like rolling on the ground after a jump, using a parachute, or even the crumple zones in cars, all designed to extend the impact time and reduce the force felt by the occupants.

The gravitational attraction between Earth and the person causes the person to accelerate towards the Earth. However, due to Earth's significantly greater mass, Earth's movement is negligible compared to that of the person. This results in the person experiencing a much greater acceleration and subsequently a larger change in momentum when they hit the ground after a jump.

Final answer:

When the charge moves 0.450 m to the right, the work done is zero. When the charge moves 0.670 m upward, the work done can be calculated as (28.0 nC * Electric Field) * 0.670 m. When the charge moves 2.60 m at an angle of 45.0° downward from the horizontal, the work done can be calculated as (28.0 nC * Electric Field * cos(45.0°)) * 2.60 m.

Explanation:

In order to calculate the work done by the electric force, we can use the equation:

Work = Force * Distance

For part (a) when the charge moves 0.450 m to the right, we need to calculate the force first using the equation:

Force = Charge * Electric Field

Given that the charge is 28.0 nC and the electric field is directed vertically upward, we should consider the vertical component of the electric field. Since the charge is moving horizontally to the right, it is perpendicular to the electric field and does not contribute to the work done by the electric force. Therefore, the work done is zero.

For part (b) when the charge moves 0.670 m upward, we can calculate the force using the same equation. Since the charge moves in the same direction as the electric field, the work done is positive. The work done can be calculated as:

Work = Force * Distance = (28.0 nC * Electric Field) * 0.670 m

For part (c) when the charge moves 2.60 m at an angle of 45.0° downward from the horizontal, we can break down the movement into horizontal and vertical components. The horizontal component does not contribute to the work done by the electric force since it is perpendicular to the electric field. The vertical component contributes to the work done. We can calculate the vertical component of the distance traveled using the angle and multiply it by the electric field to get the force. The work done can be calculated as:

Work = Force * Distance = (28.0 nC * Electric Field * cos(45.0°)) * 2.60 m

Final answer:

To find the initial velocity of a 1500 kg car that left 68-m-long skid marks, the kinetic energy (½mv²) equal to the work done by friction is used, resulting in the equation v = √(2μkgd). Plugging in the values gives the car's speed.

Explanation:

To determine how fast a 1500 kg car was traveling before it skidded to a halt on a wet road with a coefficient of kinetic friction (μk) of 0.47, leaving 68-meter-long skid marks, we can use the work-energy principle. The kinetic energy the car had before skidding must equal the work done by friction to halt the car.

Work done by friction (W) = μk × normal force (N) × distance (d)

Since the normal force (N) is equal to the weight of the car (mg) where ‘m’ is the mass of the car and ‘g’ is the acceleration due to gravity, normal force can be replaced in this equation:

W = μk × mg × d

The kinetic energy (KE) the car initially had is given by:

KE = ½ m × v2

Setting work done by friction equal to kinetic energy:

½ m × v2 = μk × mg × d

We can then solve for velocity (‘v’), which ends up being:

v = √(2 × μk × g × d)

By plugging in the values (m = 1500 kg, μk = 0.47, g = 9.8 m/s2, d = 68 m) we get the initial velocity.

Calculation:

v = √(2 × 0.47 × 9.8 m/s2 × 68 m)

After completing the calculation, we can determine the car's initial velocity that resulted in the 68-m-long skid marks on the wet road.

Answer:

The jet will fly 2400 km.

Explanation:

Given the velocity of the jet flying toward the east is 1,500 kmph toward the east.

We need to find the distance covered in 1.6 hours.

In our problem we are given speed and time, we can easily determine the distance using the following formula.

[tex]Distance=Speed\times Time[/tex]

[tex]Distance=1500\times 1.6=2400\ km[/tex]

So, the supersonic jet will travel 2400 km in 1.6 hours toward the east from its starting point.

Answer:

1920 J

Explanation:

Answer:

19.6 L

Explanation:

In this process we have an isothermal process, a transformation in which the temperature of the gas remains constant.

In an isothermal process, the work done by the ideal gas on the surrounding is:

[tex]W=nRT ln\frac{V_f}{V_i}[/tex]

where:

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas

[tex]V_i[/tex] is the initial volume of the gas

[tex]V_f[/tex] is the final volume

In this problem, we have:

n = 1.44 mol

T = 258 K is the gas temperature

[tex]V_i=14.5 L[/tex] is the initial volume of the gas

[tex]V_f=27.0 L[/tex] is the final volume

Solving the equation for W, we find the work done by the gas:

[tex]W=(1.44)(8.314)(258)ln\frac{27.0}{14.5}=1920 J[/tex]

The work done by the gas on the bullet while the bullet travels the length of a 0.54m barrel was calculated to be 9,354.80 joules, and the work done for a 1.05m barrel length was calculated to be -5,712.50 joules.

The work done by the force of the gas on the bullet can be determined by calculating the integral of the force with respect to x, from 0 to the length of the barrel. The force is a function of x, F(x) = 16,000 + 10,000x - 26,000x^2. The work done W = ∫F(x) dx, from x = 0 to x = 0.54 m.

By integrating this function we have, W = 16,000x + 5,000x^2 - 26,000x^3/3. Substituting the upper and the lower limits in, we get W = 9,354.80 joules.

For part B) we calculate work using the same formula but changing the length of the barrel to 1.05 m. The Calculated work done by the gas on the bullet, when the length of the barrel is 1.05 m, is W = -5,712.50 joules. Here, negative work signifies that the bullet is working against the direction of the force.

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To compute the work done by the gas on the bullet as it travels the length of the barrel, we use an integral to sum the work done over the length of the barrel. The force function given is integrated across the range from 0 to the length of the barrel (either 0.5400 m or 1.05 m).

The work done by a variable force in one-dimension can be computed using the formula:

Work = ∫F(x) dx

Where F(x) is given as (16,000 + 10,000x - 26,000x^2) N and the limits of the integral are from 0 to the length of the barrel. For a 0.5400 m barrel, we evaluate the integral with these limits. Similar for the second part, using 1.05 m as the upper limit.

The actual calculations would involve integrating the function for force (F), and then substituting the limits of 0 and 0.5400 m or 1.05 m in the result to calculate the work done.

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Answer:

Maximum velocity: Position C, Maximum positive acceleration: Position A.

Explanation:

Let consider that spring is compressed in the negative direction. Then, the maximum velocity occurs at position C, when spring is not compressed nor stretched. Since force in spring is of reactive nature, that is, the direction of force is opposed to the direction of movement, a maximum positive acceleration occurs at position A.

Answer:

(A). The tension in the rope that connects the boxes is 10.50 N.

(B). The value of m is 7 kg.

Explanation:

Given that,

Mass of box B = 5.00 kg

Mass of box A = m

Force = 40.0 N

Direction= 53.1°

Acceleration = 1.50 m/s²

Coefficient of kinetic friction = 0.30

(A). We need to calculate the tension in the rope that connects the boxes

Using balance equation

[tex]T=ma+m\cos\theta[/tex]

Put the value into the formula

[tex]T=5\times1.50+5.00\cos53.1[/tex]

[tex]T=10.50\ N[/tex]

(B). We need to calculate the value of m

Using formula of tension

[tex]T=ma[/tex]

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

Put the value into the formula

[tex]m=\dfrac{10.50}{1.50}[/tex]

[tex]m=7\ kg[/tex]

Hence, (A). The tension in the rope that connects the boxes is 10.50 N.

(B). The value of m is 7 kg.

The boat travels with a speed of 10m/s

Explanation:

Given-

Acceleration, a = 2m/s²

Time, t = 5s

Speed, s = ?

We know,

[tex]Acceleration = \frac{speed}{time} \\\\s = at[/tex]

[tex]s = 5 * 2m/s\\s = 10m/s[/tex]

Therefore, Boat travels with a speed of 10m/s

Answer:

8.5m/s

Explanation:

We are given that

Mass of object=m=0.50 kg

Initial velocity, u=0

Force=F=2.88 N

Time=1.48 s

a.We know that

[tex]F=ma[/tex]

Using the formula

[tex]2.88=0.50a[/tex]

[tex]a=\frac{2.88}{0.50}=5.76m/s^2[/tex]

[tex]a=\frac{v-u}{t}[/tex]

Using the formula

[tex]5.76=\frac{v-0}{1.48}[/tex]

[tex]v=5.76\times 1.48=8.5m/s[/tex]

Hence, the velocity of the object at the end of this time interval=8.5m/s

Answer:

the final velocity of the cart is 5.037m/s

Explanation:

Using the conservation of energy

[tex]T_a + V_a = T_b + V_b[/tex]

[tex]T_a = \frac{1}{2} (m_c + m_b)v_a^2[/tex]

[tex]T_a= \frac{1}{2} (3 + 0.5)(0)^2[/tex]

= 0

[tex]V_a = (m_c + m_b)gh_a[/tex]

[tex]V_a = (3 + 0.5) * 9.81 * 1.24[/tex]

[tex]= 42.918J[/tex]

[tex]T _b = \frac{1}{2} (3 + 0.5)v_b^2 \\\\ = 1.75v_b^2[/tex]

[tex]V_b = (3 + 0.5) * 9.81 * 0\\ = 0[/tex]

[tex]T_a + V_a = T_b + V_b\\ 0 + 12.918 = 1.75v_b^2 + 0\\v_b = 4.95m/s[/tex]

Using the conservation of linear momentum

[tex](m_c + m_b)v_B = m_cv_c + m_bv_b\\(3 + 0.5) * 4.95 = 3v_c - 0.5v_b\\17.33 = 3v_c - 0.5v_b\\v_b = 6v_c - 34.66 ...............(1)[/tex]

[tex]Utilizing the relative velocity relation = v_b - v_c\\-0.6 = -v_b - v_c\\v_b = 0.6 - v_c (2)[/tex]

equate (1) and (2)

[tex]6v_c - 34.66 = 0.6 - v_c\\7v_c = 35.26\\v_c = 5.037m/s[/tex]

the final velocity of the cart is 5.037m/s

Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

Now, applying the questions

So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

a=4×22/(6-4)

a=88/2

a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

Answer:

480 miles.

Explanation:

Let, S = rate on his way to the mountains.

Assume, Sgoing x time going = Sreturning x time returning

= S × 12 hours = (S + 20mph) ×8 hours

= 12 × S = 8 × S + 160.

4 × S = 160

S = 40 miles/hour

The trip took 12 hours at 40 miles per hour, so distance was:

= 12 hours × 40 mph

= 480 miles.

Answer:

[tex]3.8\times 10^{-8}\ H[/tex]

Explanation:

Given the [tex]6.7\mu F[/tex] capacitor is used in an LC circuit to produce [tex]10\ kHz[/tex] frequency.

We need to find the value of inductance required.

As we know the relation between angular frequency in [tex]rad/sec[/tex] and frequency in [tex]Hz[/tex] is.

[tex]\omega =2\times \pi\times f[/tex]

Where [tex]\omega[/tex] is angular frequency and [tex]f[/tex] is frequency.

[tex]\omega=2\times \pi\times 10\times 1000=20000\pi\ rad/sec\\\omega=62832\ rad/sec[/tex]

Also, the relation between the angular frequency, capacitance and inductance is given by.

[tex]\omega^2=\frac{1}{LC}\\\\L=\frac{1}{\omega^2C} \\\\L=\frac{1}{62832^2\times6.7\times 10^{-6} } \\\\L=\frac{1}{26451}.\\ \\L=3.8\times 10^{-8}\ H[/tex]

So, [tex]3.8\times 10^{-8}\ H[/tex] inductance will be required to produce [tex]10\ kHz[/tex].

To find the velocity of the first car after the collision, we can use the principle of conservation of momentum.

To find the velocity of the first car after the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Before the collision:

Total momentum = (mass of first car × velocity of first car) + (mass of second car × velocity of second car)

After the collision:

Total momentum = (mass of first car × velocity of first car after collision) + (mass of second car × velocity of second car after collision)

Using the given information and the principle of conservation of momentum, we can solve for the velocity of the first car after the collision.

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Given, Two boxes(Let's say A and B) are connected horizontally by a light rope. Coefficient of kinetic friction between each box is 0.30. Mass of Box B is 5.00 kg and Box A is 'm'. The force is 40.0 N in the direction [tex]53.1^{o}[/tex].

To find the tension in the rope and the mass of one of the boxes, Newton's Second Law is applied separately to each box, accounting for tension, friction, and the applied force. By resolving the applied force and calculating the friction, we can solve for tension and then for the unknown mass.

To solve for these, we must apply Newton's Second Law of Motion (F = ma) to each box separately.

Part A: Tension T in the Rope

For Box B (5.00 kg), which has the force applied:

Part B: Mass m of Box A

For Box A (with unknown mass m):

Appropriate use of Newton's laws and kinetic friction calculations will yield answers to both parts of the question.

Answer:

that the angle must be increased to maintain the minimum time of discrimination due to the increase in the speed of sound in material

Explanation:

The direction of sound is detected by the difference in time of reception of each ear, the speed of the wave is

             v = d / t

             t = d / v

In air the velocity is v = 330 m / s, let's use trigonometry

           Cos 3 = d / L

           L = d / cos 3

The difference in distance is

            Δd = d - d / cos 3 = d (1- 1 / cos3)

             t = Δd / 330

When the animal is in the water the speed of sound is

              v = 1540 m / s

So time is

             t' = Δd ’/ 1540

            t ’= Dd’ / 4.67  330

So if  t = t’  is the minimum response time, the distance must be increased

            Δd ’= 4.6 Δd

            1-1 / cos θ = 4.6 (1- 1 / cos 3) = -4.6 0.00137 = -0.00631

           1 + 0.0063 = 1 / cos θ

           1.00631 = 1 / cos θ

           Cos θ = 1 / 1.00631

           Tea = 6.5

We see that the angle must be increased to maintain the minimum time of discrimination due to the increase in the speed of sound in material

Answer:

the total number of persons between the ages of 16 and 65

Explanation:

The labour force consists of all the people who are able to work in a country or area, or all the people who work for a particular company.

The workforce of a country includes both the employed and the unemployed (labour force)

The labor force of a nation includes all employed and unemployed persons, excluding those not actively seeking work. Employed individuals have a job, while unemployed individuals are jobless but actively looking for work. The unemployment rate measures the percentage of the labor force that is unemployed.

A nation's labor force is best defined as the total number of employed and unemployed persons. Specifically, it includes all individuals who are either currently working (employed) or actively seeking employment (unemployed). The labor force does not include those who are not seeking work, such as students, stay-at-home parents, individuals with disabilities preventing them from working, or those who have taken early retirement.

The unemployment rate is a key economic indicator that represents the percentage of the labor force that is unemployed. It's important to note that to be considered unemployed, a person must be actively looking for work and available to work. This excludes people who are not actively seeking employment, who are then categorized as out of the labor force.

Lastly, the definition of 'employed' in the United States is quite broad, including those working part-time or temporarily, as well as individuals on leave but who have a job.