A tank has a shape of a cone with a radius at the top of 2 m and a height of 5 m. The tank also has a 1 m spout at the top of the tank. The tank is filled with water up to a height of 2 m. Find the work needed to pump all the water out the top of the spout. (Use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)

(a) [tex]8.13\cdot 10^{-21}[/tex]

The magnitude of the charge of one electron is

[tex]q=1.6\cdot 10^{-19}C[/tex]

Here the total amount of charge that passed through the battery pack is

Q = 1300 C

So this total charge is given by

Q = Nq

where

N is the number of electrons that has moved through the battery

Solving for N,

[tex]N=\frac{Q}{q}=\frac{1300 C}{1.6\cdot 10^{-19} C}=8.13\cdot 10^{-21}[/tex]

(b) [tex]4.16\cdot 10^5 J[/tex]

First, we can find the current through the battery, which is given by the ratio between the total charge (Q = 1300 C) and the time interval (t = 8.0 s):

[tex]I=\frac{Q}{t}=\frac{1300 C}{8.0 s}=162.5 A[/tex]

Now we can find the power, which is given by:

[tex]P=VI[/tex]

where

V = 320 V is the voltage

I = 162.5 A is the current

Subsituting,

[tex]P=(320 V)(162.5 A)=52,000 W[/tex]

And now we can find the total energy transferred, which is the product between the power and the time:

[tex]E=Pt = (52,000 W)(8.0 s)=4.16\cdot 10^5 J[/tex]

(c) 69.7 hp

Now we have to convert the power from Watt to horsepower.

We know that

1 hp = 746 W

So we can set up the following proportion:

1 hp : 746 W = x : 52,000 W

And by solving for x, we find the power in horsepower:

[tex]x=\frac{1 hp \cdot 52,000 W}{746 W}=69.7 hp[/tex]

Final answer:

The electric car moves 8.12  times [tex]10^{21}[/tex]electrons through the battery during 8 seconds of acceleration, which constitutes a 416,000 J energy transfer. The minimum horsepower rating of the car is approximately 69.7 hp.

Explanation:

To generate an accurate answer for the question posed by the student, we must apply the principles of physics regarding electric current and energy.

Part (a)

The number of electrons moved through the battery is calculated using the charge of an electron (1.60  times [tex]10^{-19}[/tex]Coulombs). For 1300 C of charge:

Number of electrons = Total charge / Charge of one electron = 1300 C / (1.60 times [tex]10^{-19}[/tex] C)

Number of electrons = 8.12 times [tex]10^{21}[/tex] electrons

Part (b)

The energy transfer is found by multiplying the total charge by the voltage of the battery:

Energy = Charge times Voltage = 1300 C  times 320 V

Energy = 416,000 J (Joules)

Part (c)

The minimum horsepower rating of the car can be found by converting the energy transfer to watts and then to horsepower:

Power (in watts) = Energy / Time = 416,000 J / 8.0 s

Power = 52,000 W

Horsepower = Power (in watts) / 746 W/hp = 52,000 W / 746 W/hp

Horsepower
= 69.7 hp (rounded to one decimal place)

Answer:

Momentum, p = 34.937 kg-m/s

Explanation:

It is given that,

Force acting on the bowling ball, F = 48.2 N

Velocity of bowling ball, v = 7.13 m/s

We have to find the momentum of the ball. Momentum is given by :

p = mv........(1)

Firstly, calculating the mass of bowling ball using second law of motion. The force acting on the ball is gravitational force and it is given by :

F = m g    (a = g)

[tex]m=\dfrac{F}{g}[/tex]

[tex]m=\dfrac{48.2\ N}{9.8\ m/s^2}[/tex]

m = 4.9 kg

Now putting the value of m in equation (1) as :

[tex]p=4.9\ kg\times 7.13\ m/s[/tex]

p = 34.937 kg-m/s

Hence, this is the required solution.

Answer:

106.1 m/s

Explanation:

The shadow of the plane is moving at the same velocity as the horizontal component of the airplane's velocity.

The horizontal component of the airplane's velocity is

[tex]v_x = v cos \theta[/tex]

where

v = 150 m/s is the velocity of the airplane

[tex]\theta=45^{\circ}[/tex] is the angle between the airplane's velocity and the horizontal

Substituting,

[tex]v_x = (150 m/s) cos 45^{\circ} = 106.1 m/s[/tex]

So, the shadow is moving at 106.1 m/s as well.

Answer:

Part a)

[tex]W = 1.2 \times 10^5 J[/tex]

Part b)

[tex]Q = 5.6 \times 10^5 J[/tex]

Explanation:

It given that player will feel exhausted when he is using his internal energy of [tex]8.0 \times 10^5 J[/tex]

PART a)

it is given that heat loss by the player is given as

[tex]Q = 6.8 \times 10^5 J[/tex]

now by first law of thermodynamics we have

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

now we have

[tex]8.0 \times 10^5 = 6.8 \times 10^5 + W[/tex]

[tex]W = 1.2 \times 10^5 J[/tex]

PART b)

It is given that another player did the work as

[tex]W = 2.4 \times 10^5 J[/tex]

now we have first law of thermodynamics

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

now we have

[tex]8.0 \times 10^5 = 2.4 \times 10^5 + Q[/tex]

[tex]Q = 5.6 \times 10^5 J[/tex]

Using the first law of thermodynamics, we find that the first player has done -1.2 x 10^5 J of work and the second player has lost 10.4 x 10^5 J of heat.

The questions posed are about applying the concept of the first law of thermodynamics in determining the amount of work done by a football player and the heat lost in the process. This law is also known as the law of energy conservation, and it can be written as ΔU = Q - W, where ΔU is the change in the system's internal energy, Q is the heat added to the system, and W is the work done by the system.

(a) The player who was dressed too lightly had to leave the game after losing 6.8 × 105 J of heat. Using the first law of thermodynamics, we can calculate the work done by the player. If the change in the internal energy before the player gets exhausted is 8.0 × 105 J (the energy used) and the heat lost is 6.8 × 105 J, then the work done (W) can be calculated as follows:
W = Q - ΔU = 6.8 × 105 J - 8.0 × 105 J = -1.2 × 105 J

(b) Another player was able to do 2.4 × 105 J of work before getting exhausted. The magnitude of the heat that he lost can then be calculated as follows:
Q = ΔU + W =  8.0 × 105 J + 2.4 × 105 J = 10.4 × 105 J

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Answer:

Internal resistance = 0.545 ohm

Explanation:

As per ohm's law we know that

[tex]V = iR[/tex]

here we know that

i = electric current = 2.75 A

V = potential difference = 1.50 Volts

now from above equation we have

[tex]1.50 = 2.75 ( R)[/tex]

now we have

[tex]R = \frac{1.50}{2.75}[/tex]

[tex]R = 0.545 ohm[/tex]

Final answer:

To find the internal resistance of a 1.50V battery shorted by a copper wire with a current of 2.75 A, you use Ohm's Law. The calculation involves dividing the total voltage by the current, yielding an internal resistance of 0.545 ohms.

Explanation:

When a 1.50V battery is shorted by a copper wire whose resistance can be ignored, the current through the copper wire is 2.75 A, you are asked to find the internal resistance of the battery. This problem can be solved using Ohm's Law, which states that the voltage (V) across a resistor is the product of the current (I) passing through it and its resistance (R), denoted as V = IR. However, when considering a battery, we must account for its internal resistance (r).

In this case, the internal resistance of the battery causes a voltage drop inside the battery, which is also governed by Ohm's Law (V = Ir).

Given the total voltage (emf of the battery, 1.50 V) and the current (2.75 A), we can rearrange the formula to solve for r: r = V/I.

Substituting the given values gives us: r = 1.50V / 2.75A = 0.545 Ω.

Therefore, the internal resistance of the battery is approximately 0.545 ohms.

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s[/tex]

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

Answer:

1.85 m

Explanation:

The horizontal velocity of the ball is

[tex]v_x = v cos \theta = (6.6 m/s) cos 39.8^{\circ}=5.1 m/s[/tex]

The horizontal distance travelled is

d = 2.7 m

And since the motion along the horizontal direction is a uniform motion, the time taken is

[tex]t= \frac{d}{v_x}=\frac{2.7 m}{5.1 m/s}=0.53 s[/tex]

The vertical position of the ball is given by

[tex]y= h + u_y t - \frac{1}{2}gt^2[/tex]

where

h = 1 m is the initial heigth

[tex]u_y = v sin \theta = (6.6 m/s) sin 39.8^{\circ}=4.2 m/s[/tex] is the initial vertical velocity

g = 9.82 m/s^2 is the acceleration due to gravity

Substituting t = 0.53 s, we find the height of the ball at this time:

[tex]y=1 m + (4.2 m/s)(0.53 s) - \frac{1}{2}(9.82 m/s^2)(0.53 s)^2=1.85 m[/tex]

Answer:

To convert kilometers per hour to meters per second, perform dimensional analysis. Remember that:

1 km = 1000 m

1 hr = 3600 seconds

Using these conversion, perform dimensional analysis:  

16.2 km/ hr (1000m/1km) (1hr/60 sec) = 4.5 m/s

The analysis basically just uses the conversion factors and canceling of units. The final answer is 4.5 m/s.  

_________________________________________________________

Correction: That should be *(1 hr/3600 sec). The answer is still 4.5 m/s.

___________________________________________________________

Hope this helps, i did the test and this answer was right, oh and brainliest, Good luck.

The wavelength of an electromagnetic wave in carbon tetrachloride with a frequency of 1.90 x 10^14 Hz and a speed of 205 x 10^8 m/s is approximately 107.89 x 10^-6 meters or 107.89 μm.

The wavelength (λ) of an electromagnetic wave can be calculated using the formula: λ = c/f

where: -

λ is the wavelength,

c is the speed of light in the medium, and

f is the frequency of the wave.

In this case, the frequency (f) is given as 1.90 x 1014 Hz, and the speed of light in carbon tetrachloride (c) is given as 205 x 108 m/s.

λ = 205 x 108 m/s / 1.90 x 1014 Hz

λ ≈ 107.89 x 10-6 m

So, the wavelength of the electromagnetic wave in carbon tetrachloride is approximately 107.89 x 10-6 meters or 107.89 μm.

With position vector

[tex]\vec r(t)=(t+1)\,\vec\imath+(t^2-1)\,\vec\jmath+2t\,vec k[/tex]

the particle then has velocity

[tex]\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\vec\imath+2t\,\vec\jmath+2\,vec k[/tex]

and acceleration

[tex]\vec a(t)=\dfrac{\mathrm d\vec v(t)}{\mathrm dt}=\dfrac{\mathrm d^2\vec r(t)}{\mathrm dt^2}=2\,\vec\jmath[/tex]

Then [tex]t=1[/tex], then particle's velocity and acceleration are, respectively,

[tex]\vec v=\vec\imath+2\vec\jmath+2\,\vec k[/tex]

and

[tex]\vec a=2\,\vec\jmath[/tex]

Explanation:

The situation described here is parabolic movement. However, as we are told the instrument is thrown upward from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

[tex]y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}[/tex]    (1)

Where:

[tex]y[/tex]  is the instrument's final position  

[tex]y_{o}=0[/tex]  is the instrument's initial position

[tex]V_{o}=15m/s[/tex] is the instrument's initial velocity

[tex]t[/tex] is the time the parabolic movement lasts

[tex]g=2.5\frac{m}{s^{2}}[/tex]  is the acceleration due to gravity at the surface of planet X.

As we know [tex]y_{o}=0[/tex]  and [tex]y=0[/tex] when the object hits the ground, equation (1) is rewritten as:

[tex]0=V_{o}.t-\frac{1}{2}g.t^{2}[/tex]    (2)

Finding [tex]t[/tex]:

[tex]0=t(V_{o}-\frac{1}{2}g.t^{2})[/tex]   (3)

[tex]t=\frac{2V_{o}}{g}[/tex]   (4)

[tex]t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}[/tex]   (5)

Finally:

[tex]t=12s[/tex]

Final answer:

Using the kinematic equation for free fall, the time it takes for the instrument to reach the point of zero velocity on planet X is 6 seconds. Since the descent takes an equal amount of time as the ascent, the total round trip time is 12 seconds.

Explanation:

To determine the time it takes for the instrument to return to its original position, we can use the kinematic equation for free fall motion under uniform acceleration, which is given by:

v = u + at

Where:

Rearranging the equation to solve for t:

t = (v - u) / a

The time it takes to reach the highest point is:

t = (0 m/s - 15 m/s) / (-2.5 m/s^2) = 6 seconds

To find the total time for the round trip, we need to double this time because the descent will take the same amount of time as the ascent:

Total time = ascent time + descent time = 6 s + 6 s = 12 seconds.

(a) [tex]1.23 m/s^2[/tex]

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

[tex]d = ut + \frac{1}{2}at^2[/tex]

Since u=0 (the block starts from rest), it becomes

[tex]d=\frac{1}{2}at^2[/tex]

So by solving the equation for a, we find the acceleration:

[tex]a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2[/tex]

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

[tex]W_p = mg sin \theta[/tex]

where

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

[tex]\theta=33.0^{\circ}[/tex] is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

[tex]F_f = - \mu mg cos \theta[/tex]

where

[tex]\mu[/tex] is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

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

Solving for [tex]\mu[/tex], we find

[tex]\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50[/tex]

(c) 12.3 N

The frictional force acting on the block is given by

[tex]F_f = \mu mg cos \theta[/tex]

where

[tex]\mu = 0.50[/tex] is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

[tex]\theta=33.0^{\circ}[/tex] is the angle of the incline

Substituting, we find

[tex]F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N[/tex]

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

[tex]v^2 - u^2 = 2ad[/tex]

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

[tex]v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s[/tex]

The acceleration of the block is 0.62 m/s². The coefficient of kinetic friction between the block and the incline is 0.048.

First, let's find the acceleration of the block. To do this, we can use the second law of motion, which states that acceleration equals the net force divided by the mass. But before we can find the net force, we need to identify the individual forces at play. The force of gravity acting on the object is 3.00 kg * 9.8 m/s² = 29.4 N. This force acts vertically downward, but since the block is on an incline, we have to resolve this force into two components: one parallel to the incline and one perpendicular to the incline. The parallel component, which is the force that actually moves the block, equals Fg * sin(33.0°) = 29.4 N * sin(33.0°) = 16.14 N. Since the block starts at rest and then speeds up, it must be accelerating. We can calculate that acceleration with the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. The problem tells us the block slides 2.00 m in 1.80 s, so we can calculate Δv using the formula Δv = Δd / Δt. Substituting the given values gives us Δv = 2.00 m / 1.80 s = 1.11 m/s. Therefore, a = Δv / Δt = 1.11 m/s / 1.80 s = 0.62 m/s².

Now let's find the coefficient of kinetic friction between the block and the incline. We know that the force of friction equals the force of gravity component perpendicular to the incline times the coefficient of kinetic friction (f = μk * Fg * cos(33.0°)), and this is equal to the force of gravity component parallel to the incline minus the force that results from the block's acceleration (f = Fg * sin(33.0°) - m*a). In other words, μk = (Fg * sin(33.0°) - m*a) / (Fg * cos(33.0°)). Substituting the given values gives us μk = (29.4 N * sin(33.0°) - 3.00 kg * 0.62 m/s²) / (29.4 N * cos(33.0°)) = 0.048.

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Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of 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{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}[/tex]

v = 2 m/s

Hence, their speed after collision is 2 m/s.

Answer:

0.733 m

Explanation:

The maximum safe intensity for human exposure is

[tex]I= 1.48 W/m^2[/tex]

Intensity is defined as the ratio between the power P and the surface irradiated A:

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

For a source emitting uniformly in all directions, the area is the surface of a sphere of radius r:

[tex]A=4 \pi r^2[/tex]

So

[tex]I=\frac{P}{4\pi r^2}[/tex]

In this case, we have a radar unit with a power of

P = 10.0 W

So we can solve the previous equation to find r, which is the distance at which a person could be considered to be safe:

[tex]r=\sqrt{\frac{P}{4\pi I}}=\sqrt{\frac{10.0 W}{4 \pi (1.48 W/m^2)}}=0.733 m[/tex]

Answer:

Angular velocity of the disk is 8.552 rad/s

Explanation:

It is given that,

Rotational kinetic energy, KE = 1280 J

The moment of inertia of the disk, I = 35 kg m²

We have to find the angular velocity of the disk. In rotational mechanics the kinetic energy of the disk is given by :

[tex]KE=\dfrac{1}{2}I\omega^2[/tex]

[tex]\omega=\sqrt{\dfrac{2KE}{I}}[/tex]

[tex]\omega=\sqrt{\dfrac{2\times 1280\ J}{35\ kgm^2}}[/tex]

[tex]\omega=8.552\ rad/s[/tex]

Hence, the angular velocity of the disk is 8.552 rad/s.

Answer:

2870 N

Explanation:

There are three forces on the mattress.  Weight of the mattress, weight of the person, and buoyancy.

∑F = ma

B - mg - Mg = 0

Buoyancy is equal to the weight of the displaced fluid.

ρVg - mg - Mg = 0

ρV - m = M

Plugging in values:

M = (1000 kg/m³) (0.75 m × 2.25 m × 0.175 m) - 2 kg

M = 293 kg

The person's weight is therefore:

Mg = 293 kg × 9.8 m/s²

Mg = 2870 N

To calculate the maximum weight (in newtons) a twin-sized air mattress can support when floating in freshwater, the buoyant force is determined by the amount of water the mattress displaces, multiplied by the density of the water. Converting displaced water weight to newtons and subtracting the weight of the mattress provides the net buoyant force, which is the maximum supportable weight.

Calculating the Buoyant Force and Supportable Weight by an Air Mattress in Freshwater

To determine how heavy a person a twin-sized air mattress can hold when placed in freshwater, we use the principle of buoyancy. Buoyancy describes the upward force exerted by a fluid that opposes the weight of an immersed object. In this case, the air mattress is the immersed object in freshwater.

The buoyant force can be found using Archimedes' principle, which states that the buoyant force is equal to the weight of the water displaced by the object. The displacement volume of the mattress can be calculated by its dimensions: 75 cm by 225 cm by 17.5 cm. However, for buoyancy calculations, we use metric units, so the dimensions should be converted to meters. The volume is thus 0.75 m x 2.25 m x 0.175 m = 0.2953125 cubic meters. The weight of the water displaced can then be calculated by multiplying this volume by the density of freshwater, which is 1000 kg/[tex]k^{3}[/tex], resulting in a displacement weight of 295.3125 kg.

To find the maximum weight the mattress can support, we convert the displacement weight to newtons (knowing that 1 kg = 9.81 N) which gives us approximately 2894.17 N. Since the mattress itself weighs 2 kg (19.62 N), the total buoyant force it can exert is the sum of the weight it can displace (2894.17 N) and its own weight. Therefore, subtracting the weight of the mattress in newtons from the total buoyant force gives us the net buoyant force, which is the maximum weight of a person it can support in newtons.

Answer:

11.3 m/s

Explanation:

Momentum is conserved:

Total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.52 kg) (11.3 m/s) + (0.52 kg) (0 m/s) = (0.52 kg) (0 m/s) + (0.52 kg) v

v = 11.3 m/s

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

[tex]I=\Delta p = m\Delta v[/tex]

where

m = 62.0 kg is the mass of the passenger

[tex]\Delta v[/tex] is the change in velocity of the car (and the passenger), which is

[tex]\Delta v = 5.30 m/s - 0 = 5.30 m/s[/tex]

So, the linear impulse experienced by the passenger is

[tex]I=(62.0 kg)(5.30 m/s)=328.6 kg m/s[/tex]

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

[tex]I=F \Delta t[/tex]

where in this case

[tex]I=328.6 kg m/s[/tex] is the linear impulse

[tex]\Delta t = 0.812 s[/tex] is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

[tex]F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N[/tex]

Answer:

0.6 kg m²

Explanation:

Angular momentum is conserved.

Iω = Iω

(1.2 kg m²) (6.0 rad/s) = I (12 rad/s)

I = 0.6 kg m²

The diver's moment of inertia in the tuck position is found to be 0.6 kg·m^2 using the conservation of angular momentum, given that no external torque acts on him.

The question relates to the concept of conservation of angular momentum, which is a principle in physics stating that if no external torque acts on a system, the total angular momentum of the system remains constant. In this problem, a diver's angular velocity increases as he changes from a relaxed position to a tucked position, indicating that his moment of inertia must decrease to conserve angular momentum because external torque is zero.

To solve for the diver's moment of inertia in the tucked position, we use the formula for conservation of angular momentum:

Therefore, the moment of inertia of the diver in the tuck position is 0.6 kg·m2.

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

Final answer:

The brightness of the 75-W 240 V bulb relative to the 75-W 120 V bulb is 50%.

Explanation:

When comparing the brightness of a 75-W lightbulb operating at 240 V in Europe to a 75-W lightbulb operating at 120 V in the United States, we can use the fact that brightness is proportional to power consumed. Since power is equal to voltage multiplied by current, we can calculate the current for each bulb using the formula P = IV. For the 75-W 240 V bulb, the current is 0.3125 A, and for the 75-W 120 V bulb, the current is 0.625 A. The brightness of the European bulb relative to the US bulb can be calculated by dividing the current of the European bulb by the current of the US bulb: 0.3125 A / 0.625 A = 0.5, or 50%.

Explanation:

a) F = GmM / r²

F = (6.67×10⁻¹¹) (500) (5.98×10²⁴) / (6.4×10⁶ + 1.5×10⁶)²

F = 3200 N

b) F = ma

3200 = 500a

a = 6.4 m/s²

c) a = v² / r

640 = v² / (6.4×10⁶ + 1.5×10⁶)

v = 7100 m/s

Answer:

A pendulum is weight suspended from a pivot so that it can swing freely

Answer:

Current, I = 2.11 A

Explanation:

It is given that,

Length of the wire, L = 3.72 m

Maximum force on the wire, F = 0.731 N

Magnetic field, B = 0.093 T

We have to find the current flowing the wire. The force acting on the wire is given by :

[tex]F=lLB\ sin\theta[/tex]

When [tex]\theta=90[/tex], F = maximum

So, [tex]F=ILB[/tex]

[tex]I=\dfrac{F}{LB}[/tex]

[tex]I=\dfrac{0.731\ N}{3.72\ m\times 0.093\ T}[/tex]

I = 2.11 A

So, the current flowing through the wire is 2.11 A. Hence, the correct option is (c).

Hostile work environment sexual harassment can be verbal, visual or physical. This statement is true.

Answer:

Option 2 is the correct answer.

Explanation:

I f the work done by a force does not depend upon the path of mass then the force is called conservative force.

Work done by frictional force depends upon path followed by mass, so frictional force is a non conservative force. But work done by gravitational force does not depend upon path followed by mass, so gravitational force is a conservative force.

Option 2 is the correct answer.

Explanation with answer:

First, in problems like this, it is always clear to draw a diagram to make sure you understand the problem.  If it is not possible to draw the diagram correctly, perhaps something is misunderstood or missing from the question.

Here, see the attached image.

Note that the rope has a tension of T that pulls both the furniture and the safe.

To find the final speed (when the safe hits the truck), we need first to find the acceleration.

The system's total mass, M = 1000+500 kg = 1500 kg

Forces acting on the system

= gravity acting on the safe less friction acting on the furniture.

= m1*g - mu*m2g

= 1000*9.81 - 0.5*500*9.81

= 7357.5 N

Acceleration, a = F/m = 7357.5 / 1500 = 4.905 m/s^2

Initial speed = 0 m/s

distance travelled, S = 3m

Let final speed = v

Kinematics equation gives

v^2-u^2 = 2aS

v^2 = 2*4.905*3 - 0^2 = 29.43 m^2/s^2

final speed, v = sqrt(29.43) = 5.4 m/s (to two significant figures.

The safe's speed when it hits the truck can be determined by considering the conversion of its initial potential energy to kinetic energy, and subtracting the work done to overcome friction.

To determine the safe's speed when it hits the truck, we apply principles of conservation of energy and account for the work done by friction forces. Initially, the gravitational potential energy of the safe is given by mgh = 1000 kg * 9.8 m/s² * 3.0 m = 29400 J. As the safe drops, its potential energy is converted to kinetic energy (0.5*mv²) while energy is also consumed to overcome the frictional force on the furniture.

The friction force is μK * N = 0.5 * 500 kg * 9.8 m/s² = 2450 N. The work done by this force over 3.0 m is 2450 N * 3.0 m = 7350 J.

As energy is conserved, the kinetic energy of the safe when it hits the truck will be the initial potential energy minus the work done on friction. So, 0.5 * 1000 kg * v² = 29400 J - 7350 J. Solving this equation will give you the speed v of the safe when it hits the truck.

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Answer:

Force exerted = 48.89 N

Explanation:

Force = Mass x Acceleration

Mass = 44 kg

Acceleration is rate of change of velocity.

Acceleration, [tex]a=\frac{2-7}{4.5}=-1.11m/s^2[/tex]

Force = Mass x Acceleration = 44 x -1.11 = -48.89 N

Force exerted = 48.89 N

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

      E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

     E = mcΔT

     725 = 0.055 x 921.096 x (T - 27.5)

     T - 27.5 = 14.31

     T = 41.81 ° C = 41.8 °C

Final temperature of the aluminum = 41.8 °C

An aluminum block of 55.0 g at an initial temperature of 27.5 °C absorbs 725 J of heat. Using the formula for heat transfer, we calculate that the final temperature of the aluminum block is approximately 36.8 °C.

This question can be answered using the formula for heat transfer Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat of aluminum and ΔT is the change in temperature. We've been given the mass, heat amount, and initial temperature. We also know that the specific heat of aluminum is 0.897 J/g°C (derived from the Table 5.1 and Table 9.1).

So the equation becomes 725 = 55 * 0.897 * (T_final -27.5). Solving for T_final, we get that the final temperature of the aluminum block is approximately 36.8 °C. Please note that this is a simplification as it doesn't take into account any heat losses to the surrounding environment.

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Answer:

Option B is the correct answer.

Explanation:

Total distance traveled = 60 + 60 = 120 m

Time taken to walk across [tex]=\frac{60}{2}=30s[/tex]

Time taken to run back [tex]=\frac{60}{6}=10s[/tex]

Total time taken = 30 + 10 = 40 s

Average speed = Total distance traveled / Total time taken

Average speed [tex]=\frac{120}{40}=3m/s[/tex]

Option B is the correct answer.

Answer:

The acceleration is [tex]11.1g\ m/s^2[/tex]

Explanation:

Given that,

Speed [tex]v= 6.50\ km/s=6.5\times10^{-3}\ m/s[/tex]

Time t = 60.0 sec

We need to calculate the acceleration

Using formula off acceleration

[tex]a = \dfrac{\Delta v}{t}[/tex]

[tex]a=\dfrac{v_{f}-v_{i}}{t}[/tex]

We know that,

Missile goes from rest

So, Initial velocity =0

Put the value into the formula

[tex]a =\dfrac{6.50\times10^{3}}{60.0}[/tex]

[tex]a=108.33\ m/s^2[/tex]

On right hand side multiplying and dividing by g = 9.8m/s²

[tex]a=108.33\times\dfrac{g}{g}[/tex]

Put the value of g

[tex]a = \dfrac{108.33}{9.8}g\ m/s^2[/tex]

[tex]a = 11.1g\ m/s^2[/tex]

Hence, The acceleration is [tex]11.1g\ m/s^2[/tex]

Answer:

- 0.86 C

Explanation:

Let the temperature of cold reservoir is T2.

T1 = 22.4 C = 295.4 K, B = 11.7

By the formula of coefficient of performance of heat pump

B = T2 / (T1 - T2)

11.7 = T2 / (295.4 - T2)

11.7 × 295.4 - 11.7 T2 = T2

T2 = 272.14 K

T2 = - 0.86 C