Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of mass m1, in meters per square second.

B) Nuclear fission

Explanation:

Nuclear power plants work by using the process of nuclear fission.

Nuclear fission occurs when a heavy, unstable radioactive nuclei decays, breaking apart into two or more lighter nuclei, more stable. In the process, several neutrons are also released, alongside with energy.

In nuclear power plants, the nucleus used for the process is the Uranium-235. When an atom of uranium-235 absorbs a slow neutron, it becomes a very unstable nucleus of uranium-236, which quickly decays into a nucleus of Barium-141, Kripton-92 and 3 neutrons.

The uranium nuclei are located in the so-called fuel rods, which are placed in a moderator (usually water). The purpose of the moderator is to slow down the neutrons emitted in the reaction: this way, these neutrons can be absorbed by other nuclei of uranium-235, causing more fission reactions to occur.

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The mass (M) of the target and the final speed (V) of the bullet and target combined after an inelastic collision can be determined by setting the total momentum before the collision equal to the total momentum after the collision and solving for M and V using the given information.

The question deals with the concept of conservation of momentum in an inelastic collision. Since the bullet and the target paper are involved in an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. This can be defined as m*v=(m+M)*V, where m and v are the mass and velocity of the bullet respectively, M is the mass of the target, and V is the final velocity of the bullet and target combined.

Given in the question is that after collision, the speed of the bullet becomes (0.516)*v, therefore the velocity V of the bullet and the target combined, the moment after the collision would be V = 0.516 * v. Solving these equations will give the required values for M and V in terms of the initial mass and velocity of the bullet.

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The question involves an inelastic collision between a bullet and a target. Using conservation of momentum, the target's mass is found to be 0.937 times the bullet's, and the target's speed immediately after collision is approximately 0.484 times the bullet's initial speed.

Given that the collision is inelastic, the quantity of momentum before and immediately after the collision is conserved. Hence, we can express this conservation of momentum as: m*v = (m+M)V. From this we get the mass M of the target as: M = [m*(1 - 0.516)]/0.516 = 0.937m which means the mass of the target is approximately 0.937 times the mass of the bullet.

Subsequently, we can solve for the speed V of the target using the above conservation of momentum giving: V = (m*v)/(m+M) = v/(1+1.064) = 0.484v approximately.

In conclusion, the mass M of the target the instant after the collision is 0.937m and the speed V of the target is 0.484v in terms of the mass m of the bullet and initial speed v of the bullet.

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Answer:

Explanation:

a)Magnitude = [tex]\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }[/tex]

84=[tex]\sqrt{(0- (-67))^{2} + (x-0)^{2} }[/tex]

x= +50.67 or -50.67 units

b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.

To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.

Magnitude = [tex]\sqrt{(0- (67))^{2} + (-130.67)^{2} }[/tex] = 146.85 units

c) The direction vector = 67/146.85 i  - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - [tex]Tan^{-1}(67/130.67)degrees[/tex] i.e 152.85 degrees from the +ve x-axis.

A) The two possibilities for the x-component are; +50.67 units or -50.67 units

B) The magnitude of the vector added to the original one is; 146.85 units

C) The direction of the vector is; θ = 207.15°

A) We are given;

Magnitude of vector = 84 units

Y-component of the vector = -67 units

We know that the formula for for 2 vectors like this in x and y direction is;

A = xi^ + yj^

Where A is the magnitude of the resultant

x is the value of the x-component

y is the value of the y-component

Thus;

A = √(x² + y²)

84 = √(x² + (-67)²)

84² = x² + 4489

7056 = x² + 4489

x = ±√(7056 - 4489)

x = ±50.67 units

B) From A above, let us take the positive value of the x-component and as such our original vector will be;

A = 50.67i^ - 67j^

We want to add another vector to this that would make the resultant to be -80 units in the x direction. Thus, A = -80i and if the new additional vector is V^, then we have;

-80i^ = (50.67i^ - 67j^) + V^

V^ = -(80 + 50.67)i^ + 67j^

V = -130.67i^ + 67j^

The magnitude of vector V is;

V = √(x² + y²)

V = √(-130.67)² + 67²)

V = 146.85 units

C) The direction of the vector V is;

θ = tan^(-1) (y/x)

θ = tan^(-1) (67/-130.67)

θ = -27.15°

Since it points entirely in the negative x axis, then the angle is;

180 - (-27.15) = 207.15°

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Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

Answer:

[tex]\vec{v} = <-11,15.08,-5>[/tex]

Explanation:

given,

location of the ball ⟨7,0,−8⟩

initial velocity of the ball ⟨-11,19,−5⟩

time = 0.4 s

speed of the ball = ?

using Momentum Principle

change in momentum = Force x time

[tex]m \vec{v} - m \vec{u}= \vec{F}\times \Delta t[/tex]

[tex]\vec{v} =\vec{u} + \dfrac{\vec{F}}{m}\times \Delta t[/tex]

Net force acting in this case will be equal to force due to gravity because air resistance is negligible.

F_net = F_g = ⟨0 ,-9.8 m , 0⟩

now,

[tex]\vec{v} = <-11,19,-5>+ \dfrac{<0 ,-9.8 m , 0>}{m}\times (0.4-0)[/tex]

[tex]\vec{v} = <-11,19,-5>+ <0 ,-3.92 , 0>[/tex]

[tex]\vec{v} = <-11,15.08,-5>[/tex]

hence, the velocity of the ball 0.4 s after being kicked is equal to [tex]\vec{v} = <-11,15.08,-5>[/tex]

The velocity of the ball 0.4 seconds after being kicked is; v = (-11, 15.08, -5) m/s

We are given;

From the Momentum Principle, we know that;

Change in momentum = Impulse

Thus;

m(v - u) = F * Δt

Divide through by m and make v the subject to get;

v = u + [(F/m)Δt]

In this question, due to the fact that air resistance is negligible, the net force acting will be equal to force due to gravity. Thus;

F_net = F_g = mg

Since acceleration due to gravity is 9.8 m/s², then;

F = (0, -9.8m, 0)

Thus;

v = (-11, 19, −5) + [((0, -9.8m, 0)/m) * (0.4 - 0)]

v = (-11, 19, −5) + (0, -3.92m, 0)

v = (-11, 15.08, -5) m/s

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Final answer:

The force on the skier parallel to the slope is found by calculating the component of the skier's weight that acts along the slope. This force is the weight multiplied by the sine of the angle of the slope, which results in 534.8 Newtons for a 103 kg skier on a 32° slope.

Explanation:

To determine the force on the skier parallel to the slope, we can make use of the component of the gravitational force along the slope. Since we are neglecting friction, the only force acting on the skier in the direction parallel to the slope is the component of the skier's weight in that direction.

The weight of the skier can be calculated by multiplying the mass (m) by the acceleration due to gravity (g), which is W = m × g. The component of the weight parallel to the slope is Wparallel = W × sin(θ), where θ is the angle of the slope. Substituting the given values, we have W = 103 kg × 9.8 m/s² = 1009.4 N. The parallel component is then 1009.4 N × sin(32°).

To find the sine of 32°, we use a calculator and get sin(32°) ≈ 0.5299. Multiplying this by the weight gives the parallel force on the skier, which is 1009.4 N × 0.5299 ≈ 534.8 N. Therefore, the force on the skier parallel to the slope is 534.8 Newtons.

Answer:

Timmy is 1 meters away from the center.

Explanation:

Given:

Distance of Suzie from center = 2 m

Weight of Suzie = 400 N

Weight of Timmy = 800 N

The see saw is balanced

To find the distance of Timmy from the center.

Solution:

Since the see saw is balanced,

So, the sum of clockwise moment = Sum of anticlockwise moments

Moment is given by:

⇒ [tex]Force\times Perpendicular\ distance\ from\ the\ center[/tex]

Let Suzies have moment in clockwise direction.

Moment of Suzie would be = [tex]400\ N \times 2\ m= 800\ Nm[/tex]

Timmy would have moment in anticlockwise direction.

Let Timmy be [tex]l[/tex] meters away from center

Moment of Timmy would be = [tex]800\ N\times l\ m=800l\ Nm[/tex]

Since, the see saw is balanced, the system is in equilibrium.

Thus, we have,

Anticlockwise Moment = Clockwise moment

[tex]800l=800[/tex]

Dividing both sides by 800.

[tex]\frac{800l}{800}=\frac{800}{800}[/tex]

[tex]l=1\ m[/tex]

Thus, Timmy is 1 meters away from the center.

Answer:

analog-to-digital

Explanation:

An analog-to-digital converter, or ADC as it is more commonly called, is a device that converts analog signals into digital signals.

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095  and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

Answer:

Explanation:

Given

Heat transfer [tex]Q=1.6\times 10^5\ J[/tex]

initial Temperature [tex]T_i=17.5^{\circ}\approx 290.5\ K[/tex]

Entropy change [tex]dS=485\ J/K[/tex]

The expression for entropy is given by

[tex]dQ=TdS[/tex]

[tex]T=\frac{dQ}{dS}[/tex]

[tex]T=\frac{1.6\times 10^5}{485}[/tex]

[tex]T=329.89\ K[/tex]

Temperature can be written as average of initial and final temperature

[tex]T=\frac{T_i+T_f}{2}[/tex]

[tex]329.89=\frac{T_f+290.5}{2}[/tex]

[tex]T_f=659.78-290.5[/tex]

[tex]T_f=369.28\ K[/tex]

Answer:

a. Memory T cells

Explanation:

Memory T cells are actually the antigen-specific T cells that remain long-term after an infection has been eliminated. These memory T cells are quickly converted into large numbers of effector T cells upon reexposure to the specific invading antigen, thus providing a rapid response to past infection that has been experienced before

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

[tex]\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\[/tex]

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

The problem involves using Charles's Law, a form of the ideal gas law, to find the final temperature of nitrogen after heating, by converting all pressures to absolute pressures and applying the law to relate initial and final states.

The student's question involves determining the final temperature of nitrogen gas in a rigid tank after heating, given initial temperature and pressures, using the ideal gas law. To solve this problem, we assume the nitrogen behaves as an ideal gas and use the relation between pressure, volume, temperature, and the number of moles of gas, which is constant since the tank is rigid.

Firstly, convert all pressures into absolute pressures by adding atmospheric pressure to the gage pressures. Then, apply the ideal gas law in the form of Charles's Law (P1/T1 = P2/T2), which relates pressure and temperature at constant volume and number of moles. To find the final temperature (T2), rearrange the equation to T2 = (P2/P1) * T1, where P2 and P1 are the final and initial absolute pressures, respectively, and T1 is the initial temperature in Kelvin.

Answer:uigu

Explanation:nvhv

The force that is experienced by the ladybug is being exerted by the can.

According to Newtons's third law of motion, action and reaction are equal and opposite. As a centripetal force is exerted on the can as it moves round the circle, the can also exerts a force back on the ladybug resting on its bottom.

Hence, the force that is experienced by the ladybug is being exerted by the can on which the ladybug sits.

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Answer:

Fr= 5400 N

Explanation:

Given that

m = 1200 kg

v= 32 m/s

μ = 0.45

Given that car is moving with constant speed it means that acceleration of the car is zero or we can say that total force on the car is zero.That is why ,only force needed to over come the friction force.

Fr= μ m g

Fr= 0.45 x 1200 x 10  N

Fr= 5400 N

That is why the total force should be 5400 N.

Answer:

Explanation:

mass, m  = 1200 kg

v = 32 m/s

coefficient of friction, μ = 0.45

As the car is moving with constant speed so the net force is zero, but the force applied by the car engine to run is the friction force.

F = μ mg = 0.45 x 1200 x 9.8 = 5292 N

Answer:

Cerebral palsy

Explanation:

Cerebral palsy - it is referred to that disorder which is related to damages that caused permanent disorder in the functioning of body parts.  it affects the proper functioning of muscles thus cause the coordination problem.

it is caused due to abnormalities in the brain that result in the coordination of the body. As it is related to abnormalities in the brain thus it also causes a problem in vision, speaking, hearing, etc

Answer:

Earth crust and specifically the continental crust.

Explanation:

If we examine the earth crust there is mostly the granite and basalt and most of the granite is present in the continental crust part which is less thicker and denser. That's why we say that the continental crust has the composition similar to that of granite.

Answer:

E- The star becomes a red giant (LATEST STAGE)

F- The surface of the star becomes brighter and cooler

C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand

A- The shell of hydrogen surrounding the star's nonburning helium core ignites.

D- The star's non burning helium core starts to contract and heat up

B- Pressure in the star's core decreases (EARLIEST STAGE)

(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.

Final answer:

Explanation of the stages in stellar evolution from subgiant to red giant.

Explanation:

Ranking the stages of stellar evolution from subgiant to red giant:

The shell of hydrogen surrounding the star's non-burning helium core ignites.

The star's non-burning helium core starts to contract and heat up.

The surface of the star becomes brighter and cooler.

Pressure from the star's hydrogen-burning shell causes the non-burning envelope to expand.

Pressure in the star's core decreases.

The star becomes a red giant.

Answer:

51.019 μT

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]

i = Current in wire = 0.203 A

N = Number of turns = 200

d = Diameter = 1 m

Magnetic field is given by

[tex]B=\dfrac{N\mu_0i}{d}\\\Rightarrow B=\dfrac{200\times 4\pi \times 10^{-7}\times 0.203}{1}\\\Rightarrow B=0.000051019\ T[/tex]

The strength of the magnetic field at this location would be 51.019 μT

The magnetic field at the given point is 51.019 μT.

It is given that the current in the coil is I = 0.203A

the diameter of the loop carrying the current is d = 2r = 1m

and the number of turns is N = 200.

The magnetic field to a current-carrying loop having N number of turns and carrying a current I with the radius of the loop being r is given by:

[tex]B=\frac{\mu_oNI}{2r} \\\\B=\frac{4\pi\times10^{-7}\times200\times0.203}{1}\\\\B=51.019\;\mu T[/tex]

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Answer: The specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]

Explanation:

As we know that,  

[tex]q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]    (1)

where,

q = heat absorbed  = 32.0 kJ = [tex]32.0\times 10^3J[/tex] J      (1kg=1000g)

[tex]m[/tex] = mass of unknown solid= 2.00 kg  = [tex]2.00\times 10^3g[/tex] (1kg=1000g)

[tex]T_{final}[/tex] = final temperature

[tex]T_{initial}[/tex] = initial temperature

[tex]\Delta T[/tex] =[tex]4.00^0C[/tex]

[tex]c[/tex] = specific heat of unknown solid = ?

Now put all the given values in equation (1), we get

[tex]32.0\times 10^3J=2.00\times 10^3g\times c\times (4.00^0C)][/tex]

[tex]c=4.00J/g^0C[/tex]

Therefore, the specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]

The scientific heat of the unknown solid will be "4.00 J/g°C".

Given values are:

Heat absorbed, q = 32.0 kJ or, [tex]32.0\times 10^3[/tex] J

Mass, m = 2.00 kg or, [tex]2.00\times 10^3[/tex] g

Rise in temperature, ΔT = 4.00°C

We know the relation,

→ q = m×c×ΔT

or,

→    = m×c×([tex]T_{final} - T_{initial}[/tex])

By substituting the values,

[tex]32.0\times 10^3=2.00\times 10^3\times c\times 4.00[/tex]

             [tex]c = 4.00[/tex] J/g°C    

Thus the above answer is appropriate.

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The correct answer is that the speed of the ball when it returns to the point of launch is less than its speed when it was initially launched, due to the effects of air resistance.

The subject of this question is Physics, specially mechanics that is a part of Physics dealing with motion and the forces that produce motion.

The correct answer is option b. 'The speed at which the ball returns to the point of launch is less than its speed when it was initially launched'. This is due to the presence of air resistance, which counteracts the force of gravity and slows the ball down both on its way up and on its way down. Therefore, the speed of the ball when it returns to the hand will be less than the speed at which it was launched.

Option a is incorrect because the net work done by gravity is zero since the ball returns to its original position. Option c is also incorrect because air resistance always does work on the ball as it changes its velocity. Lastly, option d is incorrect because the direction of air resistance changes; it is upward when the ball is rising and downward when the ball is falling.

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The correct answer is b. The ball's return speed is less than its launch speed due to the opposing forces of air resistance during its upward and downward motion.

The correct statement is b. The speed at which the ball returns to the point of launch is less than its speed when it was initially launched. This is because air resistance, a form of friction, works against the motion of the ball. When the ball is launched upward, air resistance acts downwards, slowing it down. Similarly, when the ball falls downwards, air resistance acts upwards, again slowing it down.

As a result, the return speed of the ball is less than the launch speed. It’s important to remember that, in an ideal scenario without air resistance, the return speed would be equal to the launch speed because of the conservation of energy. However, in real-world situations like this one, air resistance, a form of energy dissipation, reduces the speed.

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Answer:

4. nuclear fusion.

Explanation:

The process that helps stars generate heat energy from atomic nuclei is called nuclear fusion. Due to gravity, pressure on the hydrogen gas in the center or core of the stars are enormous causing temperature to reach about 27 million°F (15 million°C).  This hotness makes hydrogen atoms to fuse together to form helium atoms. This process is called nuclear fusion. Vast amount of energy is released which enables the star to shine.  

Answer:

What soil conditions favor the use of belled caissons?

Answer:

- where the bell can be unearthed from a solid surface.

- where the supporting stratum below the bottom of the caisson is impermeable to water movement.

What soil conditions favor piles over caissons?

Answer:

- non-cosheal soils

- subterranean water or excessive depth of bearing strata make caisson unworkable

What type of piles are especially well suited to repair or improvement of existing foundations ?

Answer:

Without hammering, minipiles or helical piles are placed which escapes much of the vibration and noise associated with traditional pile installation. for working close to existing buildings or for improving the exiting foundations where excessive vibration could damage exiting structures or noise may interfere with ongoing activities these piles are good options.

Why?

Their slenderness involves little or no displacement of the soil, thus minimizing the risk of disturbance to nearby foundations.

List and explain some cost thresholds frequently encountered in foundation design.

Answer:

building below the water table- site dewatering must occur, strengthening of slopes supper system must be done and waterproofing of the foundation all of which entails money

building near existing building - this requires underpinning(The process of reinforcing the base of an existing building or other structure underpins it.)

increase in column/wall load- building height determines the foundation depth

The answer discusses the use of belled caissons, piles, and helical piles in different soil conditions, along with common cost thresholds in foundation design.

Belles caissons are favored in soil conditions with the potential for liquefaction, as driving deep piles or piers can strengthen the soil and reduce liquefaction risk. On the other hand, piles are preferred over caissons in soil conditions that are unsuitable for drilling due to hard rock or boulders. Helical piles are especially well-suited for repairing or improving existing foundations as they can be installed quickly with minimal noise and vibration, making them ideal for retrofitting projects. Some common cost thresholds in foundation design include budgets for materials, labor, and specialized equipment, all of which can impact the overall project cost.

Answer:

4.225 J

Explanation:

Elastic Potential Energy: This is the potential energy stored in an elastic material.This also the energy required to stretch an elastic material. The S.I unit is Joules.

Mathematically it is expressed as

E = 1/2ke²....................... Equation 1

Where E =elastic potential Energy, k = spring constant, e = extension.

Given: k = 5.0 N/m, e = 3.93-2.63 =  1.3 m.

Substitute into equation 1

E = 1/2(5)(1.3)²

E = 8.45/2

E = 4.225 J.

Thus the Elastic potential Energy = 4.225 J.

Answer:

The work done by the hoop is equal to 5.529 Joules.

Explanation:

Given that,

Mass of the hoop, m = 96 kg

The speed of the center of mass, v = 0.24 m/s

To find,

The work done by the hoop.

Solution,

The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,

[tex]K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]

I is the moment of inertia, [tex]I=mr^2[/tex]

Since, [tex]\omega=\dfrac{v}{r}[/tex]

[tex]K_i=mv^2[/tex]

[tex]K_i=96\times (0.24)^2=5.529\ J[/tex]

Finally it stops, so the final energy of the hoop will be, [tex]K_f=0[/tex]

The work done by the hoop is equal to the change in kinetic energy as :

[tex]W=K_f-K_i[/tex]

W = -5.529 Joules

So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.

Answer:

You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.

Explanation:

Hi there!

The equation of velocity for an object moving in a straight line is the following:

v = v0 + a · t

Where:

v = velocity at time t.

v0 = initial velocity.

a = acceleration.

t = time.

We have to find the time at which the velocity is 100 km/h with a decceleration of 4.9 m/s² and an initial velocity of 149 km/h. Let´s first convert km/h into m/s:

149 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 41.4 m/s

100 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 27.8 m/s

Now, let´s solve the equation of velocity for the time:

v = v0 + a · t

(v - v0) / a = t

Replacing with the data:

(27.8 m/s - 41.4 m/s) / -4.9 m/s² = t

Notice that the acceleration is negative because you are slowing down.

t = 2.8 s

You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.

To get the car under the speed limit, it would take approximately 4.34 seconds.

To calculate the minimum time it takes for the car to get under the 100 km/h speed limit, we need to find the deceleration required to slow down from 149 km/h to 100 km/h.

First, let's convert the speeds to m/s. 149 km/h is equal to 41.4 m/s and 100 km/h is equal to 27.8 m/s.

Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can rearrange the equation to solve for acceleration:

a = (v² - u²) / (2s)

Plugging in the values, we get a = (27.8² - 41.4²) / (2(-30.5)).

Solving this equation, we find that the minimum deceleration required is approximately -3.13 m/s².

Finally, we can use the formula a = Δv / t to find the minimum time:

t = Δv / a = (41.4 - 27.8) / 3.13 = 4.34 seconds.

The five positions of the spaceship from left to right are based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest is [tex]5, 1, 2, 4, 3[/tex]

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another.

The gravitation force that Earth exerts on the spaceship will be:[tex]F_{ES}=(Gm_1m_E)/r^2[/tex]

Where [tex]F_{ES}[/tex] the force exerted on the spaceship by Earth [tex]m_1\\\\[/tex] is the mass of the spaceship and r is the distance between the.

[tex]F_{ES}\ \alpha\ 1/r^2[/tex]

This indicates larger the distance smaller will be the force. The correct order is [tex]5, 1, 2, 4, 3[/tex].

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The strength of the gravitational force that Earth exerts on a spaceship varies depending on the distance between them. The force is strongest when the spaceship is closest to Earth and weakest when it is closest to the Moon.

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Answer:

The heavier wagon rolls 1/2 as fast as the lighter wagon.

Explanation:

When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration  will become half of the heavier wagon in comparison with lighter one.

Answer:

0.035 N

Explanation:

Density = mass/volume

D = m/v

m = D× v ..................... Equation 1

Where D = density of the ping-pong, v = volume of the ping-pong, m = mass of the ping-pong

Note: The ping-pong is spherical in shape.

v = 4/3πr²

Where r = radius,  π = pie

d = 1.99 cm, π = 3.14

v = 4/3(1.99/2)²(3.14)

v = 4.12 cm³

Also D = 0.121 g/cm³

Therefore,

m = 0.121(4.12)

m = 0.499 g

W = mg

Where W = weight of the ping-pong

W = (0.499/1000)×9.81

W = 0.005 N.

From Archimedes principle,

Upthrust = density of water × volume of water displaced × acceleration due to gravity.

U = D'vg/1000...................... Equation 2

Note: The volume of water displaced is equal to the volume of the ping-pong.

given: v = 4.12 cm³, g = 9.81 m/s², D' = 1 g/cm³

Substitute into equation 2

U = 1(4.12)(9.81)/1000

U = 0.04 N

The force required to hold the ball completely submerged under water = U-W

= 0.04-0.005 = 0.035 N

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

Answer:

w=50 rpm

Explanation:

Given data

The mass turntable M=2kg

Diameter of the turntable d=20 cm=0.2 m

Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s

Two blocks Mass m=500 g=0.5 kg

To find

Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

[tex]I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}[/tex]

where I is moment of inertia

[tex]w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm[/tex]

The problem is a case of angular momentum conservation within the domain of rotational dynamics in physics. The turntable's initial angular momentum remains conserved despite the addition of the blocks. By accounting for the added moment of inertia from the blocks, the final angular velocity of the system can be calculated.

The subject we're discussing here comes under the physics concept of rotational dynamics particularly focusing on the conservation of angular momentum.

Before the blocks hit the turntable, we know that the turntable is rotating with an angular velocity given in RPM (revolutions per minute), which we can convert to rad/s for our calculations. So, the initial angular momentum can be represented as Lim = (moment of inertia of the system) * (initial angular velocity).

Once the blocks fall onto the turntable, they contribute to the moment of inertia of the system, while the angular momentum of the system remains conserved. Thus resulting in a decreased angular velocity. The final angular momentum can be represented as Lfm = (moment of inertia including the blocks) * (final angular velocity).

Since the initial and final angular momenta need to be equal (Lfm = Lim), we can solve the resulting equation for the final angular velocity.

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