A satellite explodes in outer space, far from any other body, sending thousands of pieces in all directions. How does the linear momentum of the satellite before the explosion compare with the total linear momentum of all the pieces after the explosion? Account for your answer.

Answers

Answer 1

Answer:

They would be the same

Explanation:

Since there's no external force, or the external net force acting on the satellite is 0. By the law of conservation of momentum, the momentum before and after the explosion must be the same.

Even if the satellite is stationary prior to the explosion, its momentum is 0. After explosion it sends many of its pieces in all direction with different non 0 momentum. The total momentum by account of their direction would sum up to be 0.

Answer 2

Final answer:

The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion.

Explanation:

The linear momentum of the satellite before the explosion is equal to the total linear momentum of all the pieces after the explosion. This is because linear momentum is conserved in an explosion. Although the distribution of momentum among the individual pieces may change, the total momentum remains the same. In this case, when the satellite explodes, the momentum of the satellite is redistributed among the thousands of pieces that are scattered in all directions.


Related Questions

You apply forces of magnitude 4.2×104N to the top and bottom surfaces of a brass cube. The forces are tangent to each surface and parallel to the sides of each surface. If the cube is 2.5 cm on a side, what is the resulting shear displacement? The shear modulus for brass is 3.5×1010Pa.

Answers

Answer: [tex]4.8(10)^{-5} m[/tex]

Explanation:

We can solve this problem by the following equation:

[tex]\eta=\frac{F.h}{A \Delta x}[/tex]

Where:

[tex]\eta=3.5(10)^{10}Pa[/tex] is the shear modulus for brass

[tex]F=4.2(10)^{4}N[/tex] is the applied force

[tex]h=2.5 cm=0.025 m[/tex] is the height of the cube

[tex]A=h^{2}=(0.025 m)^{2}=625(10)^{-6} m^{2}[/tex] is the area of each surface of the cube

[tex]\Delta x[/tex] is the shear displacement

Finding [tex]\Delta x[/tex]:

[tex]\Delta x=\frac{F.h}{A \eta}[/tex]

[tex]\Delta x=\frac{(4.2(10)^{4}N)(0.025 m)}{(625(10)^{-6} m^{2})(3.5(10)^{10}Pa)}[/tex]

Finally:

[tex]\Delta x= 4.8(10)^{-5} m[/tex]

The escape velocity from the Moon is much smaller than that from the Earth, only 2.38 km/s. At what temperature would hydrogen molecules (molar mass is equal to 2.016 g/mol) have a root-mean-square velocity vrms equal to the Moon’s escape velocity?

Answers

Final answer:

The temperature at which hydrogen molecules would have a root-mean-square velocity equal to the Moon's escape velocity is approximately 112 million Kelvin.

Explanation:

The root-mean-square speed of an ideal gas is given by the equation: vrms = sqrt(3RT/M), where R is the ideal gas constant (8.314 J mol-1 K-1), T is the temperature in Kelvin, and M is the molar mass in kg/mol. To find the temperature necessary for hydrogen molecules to achieve the Moon’s escape velocity (2.38 km/s), we first need to convert the escape velocity to m/s and the molar mass of hydrogen to kg/mol.

So, 2.38 km/s is equal to 2380 m/s, and the molar mass of hydrogen is equal 2.016 g/mol, which is 0.002016 kg/mol. Then, we solve the equation (2380 = sqrt (3RT/0.002016)) for T to get the temperature.

By resolving this equation we find that the necessary temperature would be approximately 112 million Kelvin. Therefore, for hydrogen molecules to have a root-mean-square velocity equal to the Moon's escape velocity, they must be at this extremely high temperature.

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Determine whether the following actions cause the fission reaction in the reactor to speed up or slow down.
a. speeds up fission: Adding the moderator to the reactor
b. speeds up fission: Removing the control rods from the reactor
c. slows down fission: Inserting the control rods into the reactor
d. slows down fission: Removing the moderator from the reactor
e. slows down fission: A sudden loss of primary coolant water in a pressurized water reactor

Answers

Answer:

option B and C

Explanation:

Control rod are used to regulate the nuclear reactor.

When you insert control rod in the reactor it slows down the nuclear fission inside the reactor and the energy produced in the reactor will be less.

When you remove control road from the reactor the nuclear fission increase inside the reactor and the energy production is high.

Control rod consist of boron, boron absorb the neutron which help to control the nuclear fission.

Hence, the correct answer is option B and C

What is the average power consumption in watts of an appliance that uses 5.00 kWh of energy per day? How many joules of energy does this appliance consume in a year?

Answers

Answer:

(A)  power  = 0.208 kW = 208 watts

(B)  energy = 6.6 x 10^{9} joules

Explanation:

energy consumed per day = 5 kWh

(a) find the power consumed in a day

         1 day = 24 hours

        power = \frac{energy}{time}

        power = \frac{5}{24}

          power  = 0.208 kW = 208 watts

         

(b) find the energy consumed in a year

    assuming it is not a leap year and number of days = 365 days

     1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

            energy = power x time

            energy = 208 x 31,536,000

            energy = 6.6 x 10^{9} joules

Final answer:

The average power consumption of an appliance that uses 5.00 kWh of energy per day is approximately 208.33 watts. In a year, this appliance consumes about 6.57 × 10^12 joules of energy.

Explanation:

The average power consumption of an appliance that uses 5.00 kWh of energy per day can be calculated as follows:

To convert kilowatt-hours to watts, we need to know that 1 kilowatt-hour equals to 1,000 watt-hours, and there are 24 hours in a day. Hence, 5.00 kWh is equal to 5,000 watt-hours. Therefore, the power consumption in watts is 5,000 watt-hours divided by 24 hours, yielding an average power consumption of approximately 208.33 watts.

To calculate the energy consumed in a year, we know that 1 kWh equals 3.6 million joules. Therefore, 5.00 kWh equals 18 million joules. Since there are 365 days in a year, the appliance uses 18 million joules per day times 365 days which equals 6.57 × 10^12 joules per year.

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth.
What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if
(a) its initial speed is 0.351 of the escape speed from Earth and
(b) its initial kinetic energy is 0.351 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.)
(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Answers

Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

[tex]\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0[/tex]

[tex]\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}[/tex]

[tex]\frac{0.877GMm}{R} =\frac{-GMm}{R+h}[/tex]

h = .14 R

b )

[tex]\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0[/tex]

[tex]\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}[/tex]

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

= GMm / R + 1/2 m (2GM/R)

= 0

Two 1.0 kg blocks are connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s2 by force F u .

a. What is F?
b. What is the tension at the top end of rope 1?
c. What is the tension at the bottom end of rope 1?
d. What is the tension at the top end of rope 2?

Answers

Answer:

a) [tex] F=32.0 N [/tex]

b) [tex] T_{1}=19.2 N [/tex]

c) [tex] T_{2}= 16.0 N [/tex]

d) [tex] T_{3}=3.2 N [/tex]  

Explanation:

Let's do the free body diagram per each body. In this particular case the rope has a mass so we have:

First block

[tex] F-T_{1}-m_{1}g=m_{1}a [/tex]                           (1)

Frist rope

[tex] T_{1}-T_{2}-m_{r}g=m_{r}a [/tex]                     (2)

Second block

[tex] T_{2}-T_{3}-m_{2}g=m_{2}a [/tex]                   (3)

Second rope

[tex] T_{3}-m_{r}g=m_{r}a [/tex]                               (4)

When,

T

₁ is the tension at the top of the rope 1

T

₂  is the tension at the bottom of the rope 1

T

₃ is the tension at the top of the rope 2

Now, if we add all equations from (1) to (4), we will get the value of F,

[tex] F-(m_{1}+m_{2}+2m_{r})g=(m_{1}+m_{2}+2m_{r})a[/tex]

a) Solving this equation for F:

[tex] F=(m_{1}+m_{2}+2m_{r})(a+g)[/tex]

[tex] F=(m_{1}+m_{2}+2m_{r})(a+g)[/tex]

[tex] F=32.0 N[/tex]

b) Using the equation (1), we can find T₁

[tex] T_{1}=F-m_{1}(g+a)[/tex]

[tex] T_{1}=19.2 N[/tex]

c) Let's use the equation (2) to find T₂

[tex] T_{2}=T_{1}-m_{r}(g+a) [/tex]

[tex] T_{2}= 16.0 N [/tex]

d) Using the equation (4) we can find T₃:

[tex] T_{3}=m_{r}(a+g) [/tex]  

[tex] T_{3}=3.2 N [/tex]  

Have a nice day!

A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at a speed of 6 m/s at the top. The force exerted by the track on the cart at the top of the loop is half the cart's mass, how fast is cart moving in m/s?

Answers

Answer:

The speed of cart is 5.5 m/s.

Explanation:

Given that,

Mass of cart = 100 kg

Distance = 3 m

Speed = 6 m/s

We need to calculate the speed of cart

Using relation of centripetal force and normal force

[tex]N+mg=\dfrac{mv^2}{r}[/tex]

[tex]\dfrac{m}{2}+mg=\dfrac{mv^2}{r}[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}+9.8=\dfrac{v^2}{3}[/tex]

[tex]v=\sqrt{3(\dfrac{1}{2}+9.8)}[/tex]

[tex]v=5.5\ m/s[/tex]

Hence, The speed of cart is 5.5 m/s.

Final answer:

The speed of the cart at the top of the loop is 4.24 m/s.

Explanation:

In order to find the speed of the cart at the top of the loop, we can use the principle of conservation of mechanical energy. The total mechanical energy of the cart is the sum of its gravitational potential energy and its kinetic energy. At the top of the loop, the gravitational potential energy is zero, so the total mechanical energy is equal to the kinetic energy. The kinetic energy of the cart is given by the formula KE = 1/2 * m * v^2, where m is the mass of the cart and v is its velocity.

Given that the mass of the cart is 100 kg and the radius of the loop is 3 m, we can first find the centripetal acceleration at the top of the loop using the formula a = v^2 / r. Plugging in the values, we get a = (6 m/s)^2 / 3 m = 12 m/s^2. Since the force exerted by the track on the cart at the top of the loop is half the cart's mass, we can use this force to find the net force acting on the cart. The net force is equal to the centripetal force, which is given by F = m * a. Plugging in the values, we get F = (1/2 * m) * a = (1/2 * 100 kg) * 12 m/s^2 = 600 N.

Since the net force equals the centripetal force, we can equate it to the formula F = m * v^2 / r. Plugging in the values, we get 600 N = 100 kg * v^2 / 3 m. Solving for v, we get v^2 = 600 N * 3 m / 100 kg = 18 m^2/s^2. Taking the square root of both sides, we find that v = sqrt(18 m^2/s^2) = 4.24 m/s.

Identify the true statement.A. A particle of water in a wave moves in a circular motion when viewed in cross section.B. Constructive interference of comparably sized swells tends to cancel them out and produce smooth water.C. Wave base occurs at a depth equal to one-quarter of the prevailing wavelength in an area.

Answers

Answer:

A.) A particle of water in a wave moves in a circular motion when viewed in cross section

Explanation:

Water waves are transverse waves. Waves transport energy, not water. As a wave crest passes, the water particles move in circular paths

The water particles don't move, they still end up in the place in which they began, the water particle moves up and down while the wave that passes through them moves in a perpendicular direction to the direction of the water particles moving up and down with respect to the crests and troughs of the wave (since water is a transverse wave). This will cause the wave to transverse in a circular path

The answer cannot be C because wavebase is equal to one-half the wavelength measured from still water level

Electromagnetic radiation of 5.16Ă—1016 Hz frequency is applied on a metal surface and caused electron emission. Determine the work function of the metal if the maximum kinetic energy (Ek) of the emitted electron is 4.04Ă—10-19 J.

Answers

Answer:

Work function of the metal, [tex]W_o=3.38\times 10^{-17}\ J[/tex]

Explanation:

We are given that  

Frequency of the electromagnetic radiation,  [tex]f=5.16\times 10^{16}[/tex] Hz

The maximum kinetic energy of the emitted electron, [tex]K=4.04\times 10^{-19}\ J[/tex]

We need to find the work function of the metal.

We know that the maximum kinetic energy of ejected electron

[tex]K=h\nu-w_o[/tex]

Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]

[tex]\nu[/tex] =Frequency of light source

[tex]w_o[/tex]=Work function

Substitute the values in the given formula  

Then, the work function of the metal is given by :

[tex]W_o=h\nu -K[/tex]

[tex]W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}[/tex]

[tex]W_o=3.38\times 10^{-17}\ J[/tex]

So, the work function of the metal is [tex]3.38\times 10^{-17}\ J[/tex]. Hence, this is the required solution.

Carbon dioxide makes up approximately 0.04% of Earth's atmosphere. If you collect a 2.9 L sample from the atmosphere at sea level (1.00 atm) on a warm day (27 ∘C), how many CO2 molecules are in your sample?

Answers

Answer:

number of molecules=  2.83 x 10^19

Explanation:

0.04 % means 0.04 L CO_2 in 100 L atmosphere

so for 2.9 L atmosphere CO2 vol = ( 0.04/100) x 2.9 = 0.00116

T = 27°C = 300 K , P = 1 atm ,

n = PV/RT = ( 1 x 0.00116) ÷ ( 0.08206x300) = 4.71×10^{-5}

number of molecules = 6.023 x 10^23 x 4.71×10^{-5}   = 2.83 x 10^19

An object is moving due south at a constant velocity. Then, a net force directed due west acts on the object for a short time interval, after which, the net force on the object is zero newtons. Which statement concerning the object is necessarily true?

Answers

Answer:

The south-component of its velocity is the same

Explanation:

You could say several things concerning the object.

These are some of them:

- The south-component of its velocity is the same.

- Its speed (module of the velocity) is greater now.

- The velocity has now a west-component.

- The object is now moving due west of south.

- The object did not move in a straight line.

Final answer:

After experiencing a westward net force, the object will change its velocity and once the force is removed, it will continue at a new constant velocity in a direction that combines the original southward and the newly added westward components.

Explanation:

According to Newton's first law of motion, an object at rest or moving with a constant velocity will continue to do so unless acted upon by a net force. In the scenario provided, the object initially moving due south at a constant velocity experiences a net force acting due west. This force will cause the object to accelerate in the direction of the force, which is westward, changing the object's velocity.

After the short time interval, when the net force becomes zero, the object will then move at a constant velocity in a new direction that is the resultant of the south and west velocity components.

A molecule moves down its concentration gradient using a transport protein in the plasma membrane. This is an example of

Answers

Final answer:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.

Explanation:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.

The purpose of studying a system is to identify the strengths and weaknesses of the existing system and examine current inputs, outputs, processes, security and controls, and system performance.

Answers

The purpose of system is to make system more reliable and efficient for desired work completion

Explanation:

In most of the times, the system analysts operate in a dynamic environment where change is  the necessary process . A business application, a business firm or a computer system may be the required environment. In order To rebuild a system, the key elements must be considered which we need to examine are as follows:

1. Outputs and inputs.

2. Security and Control.

3. processor

4. Environment.

5. Feedback.

6. Boundaries and interface.

An astronaut drops a hammer on the moon . It takes 1 second to hit the ground after being dropped, and it is going 1.6m/s when it lands. What is the acceleration due to gravity on thr moon?

Answers

Answer:

the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex] along downward direction

Explanation:

Here, the acceleration is constant and it is equal to acceleration due to gravity in moon. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.

v = u + at

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

[tex]v^{2} = u^{2} +2as[/tex]

where,

u = initial velocity

v = final velocity

s = displacement

a = acceleration

t = time

Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.

t = 1s

u = 0 (since the hammer is dropped)

v = -1.6m/s (since its direction is downward)

a = ?

The only equation that connects all the above quantities is

v = u + at

therefore,

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

substituting the values

a = [tex]\frac{-1.6 - 0}{1}[/tex]

a = -1.6m/[tex]s^{2}[/tex]

Thus, the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex]. The negative sign indicates that it is along downward direction.

What was the first human invention to break the sound barrier?

Answers

Answer: A Whip

Explanation:

In the chapter on "Assembling the Rooms" in The Place of Houses the houses of Andrea Palladio are discussed in terms of the way their rooms are arranged. (The Villa Rotunda which we discussed in an earlier lecture fits their descripLon perfectly.) Which of the following best describes the observaLons made there?
a.Palladio's villa plans are frequently organized around a Great Room onto which other rooms open. This room provides the basis for arranging everything else.
b.Palladio's villas generally consist of rooms loosely bunched informally along
c.Palladio's villas are always long thin buildings which enfront the street on their long side. All rooms look out on the street.

Answers

Answer:

a. Palladio's villa plans are frequently organized around a Great Room onto which other rooms open. This room provides the basis for arranging everything else.

In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter contained 75.0 g H2O initially at 23.2°C. The final temperature of the solution was 31.8°C. What was the change in enthalpy for the dissolution of this compound?

Answers

Answer:

The enthalpy for dissolution is - 305.558 J/g

Solution:

Mass of the ionic compound, m = 10.00 g

Mass of water, m' = 75.0 g

Initial temperature, T = [tex]23.2^{\circ}C[/tex]

Final Temperature, T' = [tex]31.8^{\circ}C[/tex]

Now,

To calculate the change in enthalpy:

We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]

Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g

Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]

Thus

The heat absorbed by the solution is given by:

[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]

Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]

Four particles, one at each of the four corners of a square with 2.0-m-long edges, are connected by massless rods. The masses are m1=m3=3.0 kg and m2=m4=4.0 kg. Find the moment of inertia of the system about the z axis. (the z axis runs through m2, which is at the origin, m1 is on the y axis, and m3 is on the x axis. Use the parallel-axis theorem and the result for Problem 41 to find them moment of inertia of the four-particle system about an axis that passes through the center of mass and is parallel with the z axis. Check your result by direct computation.

Answers

Answer:

Explanation:

The particles are in x-y plane with coordinates of masses as follows

m₂ at (0,0 ) m₁ at ( 0,2 ), m₄ at ( 2,2 ) and m₃ at (2,0 )

Moment of inertia about z axis

I_z = 0 + 3 x 2² + 4 x (2√2)² + 3 x 2²

= 12 + 32 + 12

= 56 kgm²

Now let us find out moment of inertia about axis through CM

According to theorem of parallel axis

I_z = I_g + m x r²  

Here m is total mass that is 14 kg and r is distance between two axis which is √2 m

56 = I_g + 14 x (√2)²

I_g  = 56 - 28

= 28 kgm²

We can directly compute  I_g  as follows

I_g = 4 x (√2)² +3 x (√2)² +4 x (√2)²+3 x (√2)²

= 8 +6 +8 +6

= 28 kgm²

So the result obtained earlier is correct.

A 4.9 kg block slides down an inclined plane that makes an angle of 27◦ with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.4 s. The acceleration of gravity is 9.81 m/s 2 . Find the coefficient of kinetic friction between the block and plane.

Answers

Answer:

μk = 0.488

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the inclined plane

fk : kinetic Friction force: parallel to the inclined plane

Calculated of the W

W= m*g

W= 4.9 kg* 9.8 m/s² = 48.02 N

x-y weight components

Wx = Wsin θ = 48.02*sin27° = 21.8 N

Wy = Wcos θ = 48.02*cos27° = 42.786 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 42.786 N

Calculated of the fk

fk = μk* FN=  μk*42.786 Equation (1)

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :

d = v₀*t+(1/2)*a*t² Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

t: time interval   (m/s)

a: acceleration ( m/s²)

Data:

d= 2.7 m

v₀ = 0

t= 5.4 s

We replace data in the formula (2)  

d = v₀*t+(1/2)*a*t²

2.7 = 0+(1/2)*a*( 5.4)²

2.7 = (14.58)*a

a = 2.7 / (14.58)

a= 0.185 m/s²

We apply the formula (1) to calculated μk:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-fk= m*a     , fk=μk*42.786 of the Equation (1)

21.8 - (42.786)*μk = (4.9)*(0.185)

21.8 -0.907= (42.786)*μk

20.89 = (42.786)*μk

μk = (20.89) / (42.786)

μk = 0.488

A 0.90-kg ball is thrown with a speed of 9.0 m/s at an upward angle of 27 ∘.A) what is its speed at its highest point, andb)how high does it go?

Answers

Answer:a)8.01 m/s

Explanation:

Given

mass of ball [tex]m=0.9 kg[/tex]

initial speed [tex]u=9 m/s[/tex]

launch angle [tex]\theta =27 [/tex]

As the projectile reaches its highest point its vertical velocity component becomes zero and there will only be horizontal component

velocity at highest Point [tex]v=u\cos \theta [/tex]

[tex]v=9\cos 27[/tex]

[tex]v=8.01 m/s[/tex]

(b)maximum height h

Maximum height is given by

[tex]H_{max}=\frac{u^2\sin^2 \theta }{2g}[/tex]

[tex]H_{max}=\frac{9^2\sin^2 (27)}{2\times 9.8}[/tex]

[tex]H_{max}=0.85 m[/tex]

Final answer:

The speed of the ball at its highest point is its horizontal velocity component, calculated using the cosine of the angle. The height reached by the ball is found using the kinematic equation for vertical motion, considering the initial vertical velocity and the acceleration due to gravity.

Explanation:

When addressing the question of a 0.90-kg ball thrown at an upward angle, we are dealing with a two-dimensional projectile motion problem in physics. This type of problem can be broken down into horizontal and vertical components. Since gravity acts only in the vertical direction, it does not affect the horizontal velocity of the projectile.

(A) Speed at the highest point: The speed of the ball at its highest point is solely the horizontal component of its initial velocity because the vertical component will be zero at that point. The horizontal velocity (
vx) can be calculated using
vx =
v cos(
θ
). So,
vx = 9.0 m/s cos(27°).

(B) Maximum height: To find the maximum height, we need to consider the vertical component of the velocity (
vy) and the acceleration due to gravity (
g), which is -9.81 m/s² (negative sign indicates the direction is opposite to the velocity). The initial vertical velocity (
vy) is
v sin(
θ
), thus
vy = 9.0 m/s sin(27°). We can then use the kinematic equation
v2 = u2 + 2as to find the maximum height (
h), where u is the initial velocity, s is the displacement, and a is the acceleration. At the highest point, the final vertical velocity (
v) is 0 m/s. Rearranging the equation to solve for
s, we get
s = v2 / (2g).

What causes the temperature to increase with height through the stratosphere

Answers

Answer: The temperature increases on the stratosphere with the altitude, given that absorption of the ultraviolet rays by the ozone.

Explanation: On the stratosphere, the water vapor and the umidity are almost nonexistents and, in view of the absorption of ultraviolet rays by the ozon, the temperature increases, reaching 35,6º Fahrenheit.  

The ozone is a unusual type of oxygen molecule. In the stratosphere, the ozone appears on a large scale and warms it up by the absorption of the ultraviolet rays energy.

Final answer:

The temperature in the stratosphere increases with altitude due to the absorption of UV radiation by ozone, which leads to exothermic reactions that generate heat, especially at higher altitudes where ozone concentration is greater.

Explanation:

The temperature increases with height in the stratosphere primarily because of the absorption of ultraviolet (UV) radiation by ozone. Ozone, a form of oxygen molecule, is very good at absorbing UV radiation from the Sun. When UV radiation is absorbed by ozone, it leads to exothermic chemical reactions which generate heat, thus raising the temperature of the surrounding atmosphere.

At higher altitudes in the stratosphere, the concentration of ozone is greater and, consequently, more UV radiation is absorbed, leading to a higher temperature in this region compared to the lower part of the stratosphere. It is this temperature inversion that distinguishes the stratosphere from the troposphere below, where the temperature normally decreases with altitude. The absorption of UV radiation and its conversion into heat makes the top of the stratosphere hotter as it is closer to the source of shortwave radiation, and the amount of absorbed shortwave radiation diminishes towards the bottom of the stratosphere.

Two charged particles each with a net charge of -q are a distance r apart. The electric force F is measured. The distance between the particles then doubles to 2r, and the amount of charge on BOTH charges doubles to -2q. Relative to the original force, what is the new electric force?

Answers

The new electric force is equal to the original force

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem, at the beginning we have

[tex]q_1 = -q\\q_2 = -q\\r=r[/tex]

So the force is

[tex]F=k\frac{(-q)(-q)}{r^2}=\frac{kq^2}{r^2}[/tex]

Later, the amount of both charges is doubled, so:

[tex]q_1' = -2q\\q_2' = -2q[/tex]

and the separation is doubled as well:

[tex]r'=2r[/tex]

So the new force is:

[tex]F'=k\frac{(-2q)(-2q)}{(2r)^2}=\frac{kq^2}{r^2}=F[/tex]

This means that the new electric force is equal to the original force.

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Suppose a force of 1.00 x 104 n is put on a 10.0 cm by 10.0 cm square. What is the pressure?

Answers

Answer:

[tex]P=1000000Pa=1MPa=1000kPa[/tex]

Explanation:

When the force that is applied is normally and uniformly distributed on a surface, the magnitude of the pressure is obtained by dividing the force applied on the corresponding area:

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

Where:

[tex]P=Pressure\hspace{3}in\hspace{3}Pa.\\F=Force\hspace{3}in\hspace{3}N\\A=Area\hspace{3}in\hspace{3}m^2[/tex]

Converting cm to m:

[tex]10cm*\frac{1m}{100cm} = 0.1m[/tex]

The area of a square is:

[tex]A=l^2=(0.1)^2=0.01m^2[/tex]

Therefore:

[tex]P=\frac{1.00\times10^4}{0.01} =1000000Pa=1MPa=1000kPa[/tex]

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9923 m at Tokyo and 0.9941 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

Answers

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

Explanation:

We know the equation

            [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

   where l is length of pendulum, g is acceleration due to gravity and T is period.

Rearranging

              [tex]g= \frac{4\pi^2l}{T^2}[/tex]

Length of pendulum in Tokyo = 0.9923 m

Length of pendulum in Cambridge = 0.9941 m

Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s

We have

                     [tex]\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998[/tex]

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

A car slams on its brakes creating an acceleration of -3.2 m/s2 it comes to a rest after traveling a distance of 210 m what was it's velocity before it began to accelerate

Answers

The initial velocity of the car is 36.6 m/s

Explanation:

The motion of the car is a uniformly accelerated motion (=constant acceleration), therefore we can apply suvat equations:

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

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem, we have:

v = 0 is the final velocity (the car comes to a stop)

[tex]a=-3.2 m/s^2[/tex] is the acceleration

s = 210 m is the displacement of the car

Solving for u, we find the initial velocity:

[tex]u=\sqrt{v^2-2as}=\sqrt{0-(2)(-3.2)(210)}=36.6 m/s[/tex]

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Consider the following electron configurations to answer the question:
(i) 1s2 2s2 2p6 3s1
(ii) 1s2 2s2 2p6 3s2
(iii) 1s2 2s2 2p6 3s2 3p1
(iv) 1s2 2s2 2p6 3s2 3p4
(v) 1s2 2s2 2p6 3s2 3p5
The electron configuration of the atom that is expected to have a positive electron affinity is ________.

Answers

Answer:

(ii) 1s2 2s2 2p6 3s2

Explanation:

Electron Affinity is the energy change that occur when an atom gains an electron.

                                  X₍₉₎ + e⁻  →  X⁻           ΔE = Eea

ΔE is change in energy

Eea is electron affinity

Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:

1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.

2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.

3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.

Applying these consideration to the elements given in the question:

(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)

(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)

(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)

(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)

(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)

The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².

Final answer:

An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Typically, these atoms tend to have half-filled or filled subshell configurations. In the given examples, the atom with electron configuration 1s2 2s2 2p6 3s2 3p1 is likely to have a positive electron affinity.

Explanation:

The electron affinity of an atom is a measure of the energy change when an electron is added to a neutral atom to form a negative ion. An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Usually, the atom would tend to have a half-filled (e.g. 3p3) or filled (e.g. 3p6) subshell configuration.

If we consider the given electron configurations: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5, it appears that (iii) 1s2 2s2 2p6 3s2 3p1 should have a positive electron affinity. This atom is a Potassium atom (K). It has a positive electron affinity because it is just one electron away from obtaining a full orbit shell (in the 3s and 3p orbitals) which would provide a more stable electron configuration.

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The masses of the blocks are m1 = 16.0 kg and m2 = 12.0 kg, the mass of the pulley is M = 5.00 kg, and the radius of the pulley is R = 0.300 m. Block m2 is initially on the floor, and block m1 is initially 4.60 m above the floor when it is released from rest. The pulley's axis has negligible friction. The mass of the string is small enough to be ignored, and the string does not slip on the pulley, nor does it stretch.

a. How much time (in s) does it take block m1 to hit the floor after being released?
b. How would your answer to part (a) change if the mass of the pulley were neglected? (Enter the time, in seconds, it takes block m1 to hit the floor if the mass of the pulley were neglected.)

Answers

Answer:

a)Time taken will be 2.783 s

b)Time taken will be 2.564 s

Explanation:

a)Since the pulley has mass ,

It will have a MOMENT OF INERTIA . in other terms, whn the string slides upon it, it will produce a torque ( due to the tension) and thus it will make the pulley roll.The string soesn't get slackened. Thus the acceleration along the string must be constant - which is the string constraint. the FBD's of the bodies are attached. from them ,

[tex]m_{1}g-T_{1}= m_{1}a//T_{2}-m_{2}g=m_{2}a[/tex] ------3

Since the string doesn't slip, the acceleration of the pulley at the end point of contact of the string must be equal to [tex]a[/tex]or,

             αR = a ;                    ------1

writing the torque equation about COM of the pulley , we get

[tex](T_{1} - T_{2})*R=m*R^{2}*(alpha)=5*R^{2}*\frac{a}{R}[/tex] ------2

solving these we get ,

[tex]a=\frac{4g}{33}[/tex]

a)time taken :

[tex]s=ut+\frac{1}{2} at^{2}\\u=0\\a=\frac{4g}{33} \\s=4.6\\4.6=\frac{1}{2} * \frac{4*9.8}{33} *t^{2}\\t= 2.783 sec[/tex]

ANS : 2.783 sec

b)

In case of B, mass is zero.

Thus, there is no rotation of the pulley. this is equivalent to a normal 1 Dimension motion question

equations are:

[tex]m_{1}g-T_{1}=m_{1}a\\T_{1}-m_{2}g=m_{2}a\\a= \frac{m_{1}-m_{2}}{m_{1}+m_{2}}g\\a=\frac{4g}{28} \\a=\frac{g}{7} \\a=1.4ms^{-2}[/tex]

Thus time t will be ,

[tex]s=\frac{1}{2} at^{2}\\4.6=\frac{1}{2}*1.4*t^{2}\\ t=2.564 sec[/tex]

ANS : 2.564 sec

The time it takes the block to reach the ground can be found by making

use of the law of conservation of energy.

a. The time it takes the the block m₁ to hit the floor is approximately 2.596 seconds.

b. The time it takes the the block m₁ to hit the floor,  if the mass of the pulley were neglected is approximately 2.562 seconds.

Reasons:

Given parameters are;

Mass of block m₁ = 16.0 kg

Mass of block m₂ = 12.0 kg

Mass of the pulley, M = 5.00 kg

By conservation of energy, we have;

m₁g·h - m₂·g·h = 0.5×(m₁ + m₂)·v² + 0.5·I·ω²

[tex]\omega = \dfrac{v}{R}[/tex]

[tex]Moment \ of \ inertia\ of \ pulley, I = \dfrac{1}{2} \cdot M \cdot R^2[/tex]

Therefore;

[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2 + 0.5 \cdot I \cdot \dfrac{v}{R}[/tex]

Which gives;

(16 - 12)×9.81×4.6 = 0.5×(16+12)×v² + 0.5×(0.5×5×0.3²)× [tex]\left(\dfrac{v}{0.3} \right)^2[/tex]

Solving gives, v ≈ 21.93 m/s

We have;

v ≈ 3.544

v² = 2·a·h

[tex]a = \dfrac{v^2}{2 \times h}[/tex]

Which gives;

[tex]a = \dfrac{3.544^2}{2 \times 4.6} \approx 1.365[/tex]

v = a×t

[tex]t = \dfrac{v}{a} = \dfrac{3.544}{1.365} \approx 2.596[/tex]

The time it takes the the block m₁ to hit the floor, t ≈ 2.596 seconds

b. When the mass of the pulley is neglected, we have;

[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2[/tex]

(16 - 12)×9.81×4.6 = 0.5×(16+12)×v²

180.504 = 14·v²

[tex]v = \sqrt{\dfrac{180.504}{14} } \approx 3.591[/tex]

[tex]a = \dfrac{3.591^2}{2 \times 4.6} \approx 1.401[/tex]

[tex]t = \dfrac{v}{a} = \dfrac{3.591}{1.401} \approx 2.562[/tex]

The time it takes the the block m₁ to hit the floor,  if the mass of the pulley were neglected, t ≈ 2.562 seconds.

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An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 65 rpm to 90 rpm in 10 s .

Answers

Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

[tex]a_{t}=r\alpha[/tex]

[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}[/tex]

[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}[/tex]

[tex]a_{t}=0.0301\ m/s^2[/tex]

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

A 0.150-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.510 m apart. If the coefficient of kinetic friction between the rod and rails is 0.160, what vertical magnetic field is required to keep the rod moving at a constant speed?

Answers

Answer:

B = 0.0307 T = 30.74 mT

Explanation:

Given

m = 0.150 kg

I = 15.0 A

d = 0.510 m

μk = 0.16

B = ?

Balancing the forces on the rod in the j direction

N - m*g = 0   ⇒   N = m*g

and in the i direction

I*d*B - μk*N = 0     ⇒      B = μk*N / (I*d)

⇒      B = μk*m*g / (I*d)

⇒      B = (0.16)*(0.150 kg)*(9.8 m/s²) / (15.0 A*0.510 m)

⇒      B = 0.0307 T = 30.74 mT

Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A is under a tension of 120.00 N. String B is under a tension of 130.00 N. They are each plucked and produce sound at the n=10 mode. What is the beat frequency?

Answers

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × [tex]\frac{ \lambda }{2}[/tex]      ........1

so  λ = [tex]\frac{2L}{10}[/tex]  

and velocity is express as

V = [tex]\sqrt{\frac{T}{\mu } }[/tex]    .................2

so

frequency for string A = f1 = [tex]\frac{V1}{\lambda}[/tex]

f1 = [tex]\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}[/tex]

f1 = [tex]\frac{10}{2L} \sqrt{\frac{T1}{\mu } }[/tex]      

and

f2 = [tex]\frac{10}{2L} \sqrt{\frac{T2}{\mu } }[/tex]

so

beat frequency is = f2 - f1

put here value

beat frequency = [tex]\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}[/tex]  - [tex]\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }[/tex]

beat frequency = 13.87 Hz

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